4x2 + — b2 — 2ab a = 2x, b€¦ · rd sharma solutions class 9 algebraic identities ex 4.1 rd...
TRANSCRIPT
RD Sharma Solutions Class 9 Algebraic Identities Ex 4.1
RD Sharma Solutions Class 9 Chapter 4 Ex 4.1
1) Evaluate each of the following using identities:
0)( 2 z - £ ) 2
Solution:
Given,
( 2 x - ± ) 2 = (2x)2 + (± ) 2 - 2 * 2 x * ±
(2x — - j ) 2 = 4x2 + — 4 [•.• (a — b)2 = a2 + b2 — 2ab ]
Where, a = 2x, b = ^
.-. (2a: — -i-)2 = 4a;2 + — 4
00 (2x+y) (2x-y)
Solution:
Given, (2x+y) (2x-y)
= (2a:)2 - (y ) 2 [•.■ (a + b)(a — b) = a 2 - b2]
= 4a:2 - y2
(2a: + y){2x - y) = 4a:2 - i/2
(Hi) (a 2b — ab2)2
Solution:
Given, (a 2b — ab2)2
= (a2b)2 + (ab2)2 — 2 * a 2b * ab2 [v (a — b)2 = a 2 + b2 — 2a6]
Where, a = a 2b, b = ab2
— a462 + b4a2 - 2a 3b3
(a 2b — ab2)2 — a4&2 + b4a2 — 2a363
(iv) (a-0.1) (a+0.1)
Solution:
Given, (a-0.1) (a+0.1)
= a 2 — (0.1)2 ['.• (a + b)(a — b ) = a 2 — b2]
Where, a = a and b = 0.1
= a 2 - 0.01
(o - 0.1)(o + 0.1) = o2 - 0.01
(v) (1.5a;2 - 0.3j/2)(1.5a:2 + 0.3y 2)
Solution:
Given, (1.5a;2 — 0.3y2)(1.5a:2 + 0.3y 2)
= (1.5a;2) 2 — (0.3 y 2)2 [•_• (a + b)(a — b) = a 2 — 62]
Where, a = 1.5a:2, b = 0.3y2
= 2.25a:4 - 0.09y4
.-. (1.5a:2 - 0.3y2)(1.5a:2 + 0.3i/2) = 2.25a:4 - 0.09y4
2) Evaluate each of the following using identities:
0) (399)2
Solution:
We have,
3992 = (400-1 )2
= (400)2+(1 )2 - 2x400x1 [ (a-b)2 = a2+ b2-2ab ]
Where, a = 400 and b = 1
= 160000 + 1 -8000
= 159201
Therefore, (399)2 = 159201.
(ii) (0.98f
Solution:
We have,
(0.98)2 = (1-0.02)2
= (1)2+(0.02)2- 2x1x0.02
= 1 + 0.0004 - 0.04 [ Where, a=1 and b=0.02 ]
= 1.0004-0.04
= 0.9604
Therefore, (0.98)2 = 0.9604
(Hi) 991x1009
Solution:
We have,
991x1009
= (1000-9) (1000+9)
= (1000)2 - (9)2 [ (a+b) (a-b) = a2 - b2 ]
= 1000000-81 [ Where a=1000 and b=9]
= 999919
Therefore, 991 xl 009 = 999919
(iv) 117x83
Solution:
We have,
117x83
= (100+17) (100-17)
= (100)2 - (1 i f [ (a+b) (a-b) = a2 - b2 ]
= 10000-289 [ Where a=100 and b=17 ]
= 9711
Therefore, 117x83 = 9711
3) Simplify each of the following:
0) 175x 175+2x 175x25+25x25
Solution:
We have,
175 x 175 +2 x 175 x 25 + 25 x 25 = (175)2 + 2 (175) (25) + (25)2
= (175+25)2 [ a2+ b2+2ab = (a+b)2 ]
= (200)2 [ Where a=175 and b=25 ]
= 40000
Therefore, 175 x 175 +2 x 175 x 25 + 25 x 25 = 40000.
00 322x322 - 2x322x22+22x22
Solution:
We have,
322 x 322 - 2 x 322 x 22 + 22 x 22
= (322-22)2 [ a2+ b2-2ab = (a-b)2 ]
= (300)2 [ Where a=322 and b=22 ]
= 90000
Therefore, 322 x 322 - 2 x 322 x 22 + 22 x 22= 90000.
OH) 0.76x0.76+2x0.76x0.24 + 0.24x0.24
Solution:
We have,
0.76 x 0.76 + 2 x 0.76 x 0.24 + 0.24 x 0.24
= (0.76+0.24)2 [ a2+ b2+2ab = (a+b)2 ]
= (1,00)2 [ Where a=0.76 and b=0.24 ]
= 1
Therefore, 0.76 x 0.76 + 2 x 0.76 x 0.24 + 0.24 x 0.24 = 1.
0v) 7.83*7 .83-1 .17*1 .176.66
Solution:
We have.
7.83*7.83—1.17*1.176.66
_ (7 .83+ 1.17)(7.83—1.17) 6(66
_ (9.00)(6.66)(6.66)
[•.• (o - b)2 = ( a + b)(a — 6)]
. 7 .83*7 .83-1 .17*1 .17 _ q
6.66 y
4) I f x + ^ = 11, find the value o f x2 + 4
Solution:
We have, x + -j- = 11
Now, (a: + ± )2 = a:2 + ( )2 + 2 * x * j^ (x + ±)2 = x 2 + J _ + 2
=+(11)2 = x2 + 4 + 2 a: + ^ = 11]
=> 121 = x2 + 4 + 2 => a;2 + 4 = 119
X 25) I f x — — = —1, f in d the value o f x2 + 4 Solution:
We have, x — — = — 1X
Now, (x — |-)2 = a;2 + (-i-)2 — 2 * x *^(x-^)2 = x2 + 4 - 2^(-l)2 = a:2 + i - 2 [vx-± = -l]=>2 + 1 = a:2+ 4
X 2
6) I f x + = \/h, f in d the value o f x2 + 4 and x4 + 4 Solution:
We have,
(* + )2 = x2 + (i)2 + 2 * x * ±
^ ( X + ± f = X2 + ± + 2
=> (-s/5)2 = x2 + 4 + 2 [v x + = \/5]+►5 = a:2+ 4 + 2
X I^ * 2 + ^ = 3 ...(1)
Now, (a:2 + 4)2 = *4 + 4 + 2*a:2*4^ ( a : 2 + 4 ) 2 = x4 + 4 + 2
=>9 = x4 + 4 + 2 [••• a:2 + ^ = 3]
^ * 4 + V = 7
Hence, x2 + 4 = 3; x4 + 4 = 7-X £ X 4
7) I f x2 + 4 = 66, f in d the value o f x — ^
Solution:
We have,
(x-i)2 = x2 + (i)2-2*x*|^(x_i)2 = x2 + 4 - 2=> (x - |)2 = 66 - 2 [+ x2 + 4 = 66]
=► (* - | )2 = 64
= K * - i ) a = (± 8 )2
=► * - * = ± 8
8) I f x 2 + = 79, find the value o f x + ^
Solution:
We have,
( ! E + I ) 2 = a;2 + ( I ) 2 + 2 * a;* i
^ {x + 1 )2 = x 2 + ± + 2
^ ( x + |)2 = 79 + 2 [v*a + = 7 9 ]
= > (* + ± ) a = 81
=>(* + £ )a = (±9)a
=> * + i = ±9
9) if9x*+25f= 181 andxy=-6, find the value of 3x+5y.
Solution:
We have,
(3x + 5y)2 = (3x)2 + (5y)2 + 2*3x*5y
=>(3x + 5y)2 = 9x2 + 25y2 + 30xy
= 181+ 30(-6) [ Since, 9x2 + 25y2 = 181 and xy = -6 ]
=>(3x+5y)2 = 1
=>(3a; + by)2 = (±1)2
=>3* + by = ±1
10) lf2x+3y=8 and xy=Z find the value o f + 9f.
Solution:
We have,
(2x + 3y)2 = (2x)2 + (3y)2 + 2*2x*3y
=> (2x + 3y)2 = 4x2 + 9y2 + 12xy [Since, 2x + 3y = 8 and xy = 24 ]
=> (8)2 = 4x2 + 9y2 + 24
=> 64 - 24 = 4x2 + 9y2
=> 4x2 + 9y2 = 40
11) lf3x-7y=10 and xy=-1, find the value of Sx2+49y*.
Solution:
We have,
(2 - 7y)2 = (3x)2 + (-7y)2 - 2*3x*7y
=> (3x - 7y)2 = 9x2 + 49y2 - 42xy [Since, 3x - 7y = 10 and xy = -1 ]
=> (10)2 = 9x2 + 49y2 + 42
=>100- 42 = 9x2 + 49y2
=> 9x2 + 49y2 = 58
12) Simplify each of the following products:
0) - 36)(36 + \ a ) { \ a 2 + 9b2)
Solution:
(| a -3 6 )(3 6 + ± a )(| a 2 + 962)
=> K {« )2 - (3&)2)] \ { \a 2 + 9 ft2)] [•.• (a + b){a - b ) = a 2 - b2)}
^ [ \ a 2 -9t i2) ] [ \ a 2 + 9b?] [v (ab)2 = a2b2)\
= [(| «2)2 - (9&2)2] [••■ (a + 6 )(o - b ) = a2 - b2)\
= ^ a 4 -8 1 6 4
(\a - 36)(36+ f a ) ( {a 2 + 962) = ^ a 4 - 8164
0 0 { m + ± Y ( m - f )
Solution:
We have,
( m + y ) 3 ( m - y )
= ( ™ + f ) ( m + f ) ( m + f ) ( m - f )
= ( m + y ) 2 ( m 2 — ( y ) 2) [•.■ (a + 6)(a + 6) = (a + 6)2 an d (a + b)(a — 6) = a2 — 62]
= (m + y ) 2 K -( m + ? ) 3 ( m - ^ ) = ( m + ^ ) 2 ( m 2 - £ )
W ( f - f ) ( f - f ) - a;2 + 2a:
Solution:
We have,
(t - ! ) ( f - f ) - * 2 + 2*
= > - ( ! - ! ) (f - f ) - * 2 + 2*
=► - ( I - f ) 2 - X 2 + 2z [•.• (a - 6)(a - 6) = (a - 6)2]
^ - [ ( f )2 + ( ! ) 2 - 2(f ) (1 ) ] - * 2 + 2*
^ - ( ^ + T - f ) - ^ + 23:
^ - T - ^ + f + 2 x ~ i5x2 1 2 a ;____£
^ 4 ^ 5 25
• _ 2 \ /2 _ x\■ * ^2 5 M 5 2 * + 2z = ^ _4_
25
W ( * 2 + a; — 2) (a;2 — z + 2)
Solution:
( z 2 + x — 2) (z 2 — z + 2)
[(z ) 2 + (z - 2 ) ] [(z 2 - ( z + 2)]
=>(z2) 2 — (z — 2)2 [ (a - b) (a + b) = a2 - b2]
=>z4 — ( z 2 + 4 — 4 z) [y (a — 6)2 = a2 + 612 — 2a6]
=>z4 — z 2 + 4z — 4
(z2 + z — 2) (z2 — z + 2) = z4 — z2 + 4z — 4
(z3 — 3z — z) (z2 — 3z + 1)
Solution:
We have,
(z3 — 3z — z) (z2 — 3z + 1)
=>z(z2 — 3z — 1) (z2 — 3z + 1)
=>z[(z2 — 3z)2 — ( l )2] [•/ (a + 6)(a — 6) = a2 — 62]
=>z[(z2)2 + (—3z)2 — 2(3z)(z2) — 1]
=>z[z4 + 9z2 — 6z3 — 1]
=>z5 — 6z4 + 9z3 — z
(z 3 — 3z — z ) ( z 2 — 3z + 1) = z 5 — 6z4 + 9z3 — z
(vl) (2 z4 — 4 z2 + 1) (2 z4 — 4 z2 — 1)
Solution:
We have,
(2 z4 - 4 z2 + 1) (2 z4 - 4 z2 - 1)
=>[(2z4 - 4 z2) 2 - ( l ) 2] ['.• (a + 6)(0 - 6) = a2 - 9 ]
=>[(2£C4)2 + (4a:2)2 - 2(2a;4)(4a:2) - 1]
=>4a:8 — 16a:6 + 16a:4 — 1 [v (a — 6)2 = a2 + b2 — 2ab]
(2a:4 — 4a:2 + 1) (2a;4 — 4a:2 — 1) = 4a:8 — 16a:6 + 16a:4 — 1
13) Prove that a2 + b2 + c2 — ab — bc — ca is always non-negative for all values of a, b and c.
Solution:
We have,
a2 + b2 + C2 — ab — be — ca
Multiply and divide by '2'
= I [o2 + ft2 + c2 — ab — be — co]
= \ [2a2 + 2b2 + 2c2 - 2ab - 26c - 2co]
= -| [a2 + a2 + b2 + b2 + c2 + c2 — 2a6 — 26c — 2ca]
= [(a2 + 62 — 2a6) + (a2 + c2 — 2ca) + (62 + c2 — 26c)]
= 2 K « - ^)2 + (6 — c)2 + (c — a)2] [y (a — 6)2 = a2 + b2 — 2ab\
a2 + 62 + c2 — 06 — 6c — ca > 0
Hence, a2 + b2 + c2 — a6 — 6c — ca > 0 is always non-negative for all values of a, b and c.
RD Sharma Solutions Class 9 Algebraic Identities Ex 4.2
RD Sharma Solutions Class 9 Chapter 4 Ex 4.2
Question 1: Write the following in the expand form:
(i) : (a + 26 + c)2
(ii) : (2a - 36 - c)2
(iii) : (-3a; + y + z)2
(iv) : (m + 2n — 5p)2
(v) : (2 + x - 2y f
(vi) : (a2 + 62 + c2) 2
(vii) : (ab + bc + ca) 2
(viii): ( f + f + f )2
(* > (£ + £ + ^ ) 2
(x): (x + 2y + 4z)2
W): ( 2 x - y + z)2
(xii): (~2x + 3 y + 2 z)2
Solution 1 (i):
We have,
(a + 2b c f — a2 + (26)2 + c2 2a(26) + 2ac + 2(26)c
[•.• (x + y + z f = x 2 + y 2 + z2 + 2 xy + 2yz + 2 xz]
(a + 26 + c)2 = a2 + 4b2 + c2 + 4ab + 2ac + 4be
Solution 1 (ii):
We have,
(2a - 36 - c)2 = [(2a) + (-36) + (-c)]2
(2a)2 + (—36)2 + ( - c )2 + 2(2a)(—36) + 2 (-36 )(-c ) + 2(2a)(-c)
[•.• (x + y + z)2 = x 2 + y 2 + z2 + 2 xy + 2 yz + 2 xz]
4a2 + 962 + c2 — 12a6 + 66c — 4ca
••• (2a-3b-c)A{2}=4xA{2}+9yA{2}+cA{2}-12ab+6bc-4ca
Solution 1 (iii):
We have,
(—3® + y + z)2 = [(-3®)2 + y2 + z2 + 2(-3®)j/ + 2yz + 2 (-3 x)z
[•.• (x + y + z)2 = x 2 + y 2 + z2 + 2 xy + 2 yz + 2 xz]
9®2 + y2 + z 2 — 6xy + 2 yz — 6 xz
(—3® + y + z)2 = 9®2 + j/2 + z2 — 6 xy + 2 xy — 6 xy
Solution 1 (iv):
We have,
(m + 2n — 5p)2 = m 2 + (2n )2 + ( — 5p)2 + 2m x 2n + (2 x 2n x — 5p) + 2m x — 5p
[v (® + y + z f = x2 + y2 + z2 + 2xy + 2yz + 2xz]
(m + 2n — 5p)2 = m 2 + 4n2 + 25p2 + 4mn — 20np — 10pm
Solution 1 (v):
We have,
(2 + x - 2y)2 = 22 + x 2 + (~2y)2 + 2(2)(®) + 2 (® )(-2 y ) + 2 (2 )(-2 y )
[y (® + y + z)2 = x 2 + y2 + z2 + 2xy + 2yz + 2xz]
= 4 + x 2 + 4 y2 + 4® — 4xy — 8 y
(2 + x — 2 y)2 = 4 + x 2 + 4 y2 + 4® — 4xy — 8 y
Solution 1 (vi):
We have,
(a2 + 62 + c2)2 = (a2)2 + (62)2 + (c2)2 + 2 a262 + 2b2 c2 + 2a2c2
[■.■ (® + y + z)2 = x 2 + y2 + z2 + 2 xy + 2 yz + 2 xz]
(a2 + 62 + c2)2 = a4 + 64 + c4 + 2a262 + 26V + 2c2a2
Solution 1 (vii):
We have,
(ab + be + ca)2 = (a&)2 + (6c)2 + (ca)2 + 2(a6)(6c) + 2(&c)(ca) + 2(a6)(ca)
[y (x + y + z)2 = x 2 + y 2 + z2 + 2xy + 2yz + 2xz]
= a2b2 + b2c2 + (?a2 + 2(ac)62 + 2(a6)(c)2 + 2(6c)(a)2
(ab + be + ca)2 = a262 + &2c2 + c2a2 + 2ac62 + 2abc2 + 2bca2
Solution 1 (viii):
We have.
( f + ! + 7 )2 = ( f )2 + ( ! ) 2 + ( I )2 + 2 f 7 + 2f 7 + 2^ f['.• (a: + y + z)2 = a;2 + y2 + z2 + 2xy + 2 yz + 2 xz\
■ • • ( f + f + i )2 = ( S ) + ( S ) + ( ^ ) + 2 f + 2 f + 2fSolution 1 (ix):
We have,
( e + = + ^ ) 2 = ( £ ) 2 + ( i r ) 2 + ( ^ ) 2 + 2 ( £ ) ( £ ) + 2( £ ) ( £ ) + 2( e ) ( 3 )
[■.■ (a-, + y + z)2 = x 2 + y 2 + z2 + 2xy + 2 yz + 2xz]( «_ + A _£.)2 = (_aL\ , ( J L ) + (_*!_) _|— 2— |— 2— |— 2_t be ' co ~ ab > 'P62 1 ' Kcla2 ' ' 'aW ' ~ a? ' & ' c2Solution 1 (x):
We have,
(x + 2y + 4z)2 = x2 + (2y)2 + (4z)2 + 2x x 2y + 2 x 2y x 4z + 2x x Az
[•.• (x + y + z)2 = x 2 + y 2 + z2 + 2xy + 2 yz + 2 xz]
(a; + 2 y + 4z)2 = x2 + 4 y2 + 16 z 2 + 4 xy + 16 yz + 8 xz
Solution 1 (xi):
We have,
(2 x - y + z)2 = (2a:)2 + { -y )2 + (z)2 + 2(2 x ){ -y ) + 2(-y)(z) + 2(2 x)(z)
[■.• (a:+ y + z)2 = x 2 + y2 + z2 + 2 xy + 2 yz + 2xz]
(2x - y + z)2 = Ax2 + y 2 + z 2 — 4a:y — 2yz + 4xz
Solution 1 (xii):
We have,
(-2a : + 3 y + 2 z)2 = (-2 a :)2 + (3y)2 + (2*)2 + 2(— 2as)(3y) + 2(3y)(2z) + 2 (-2® )(2z)
[■.■ (x + y + z)2 = x 2 + y 2 + z2 + 2 xy + 2 yz + 2 xz]
( — 4® + 6 y + 4 z)2 = 4®2 + 9 y2 + 4 z 2 — 12xy + 12 yz — 8 xz
Question 2: Use algebraic identities to expand the following algebraic equations.
Q2.1:(a + b + c)2 + (a — b + c)2
Ans: We have,
(a + b + c)2 + (a — b + c)2 =(a2 + 62 + c2 + 2ab + 2bc + 2 co) + (a2 + (~b)2 + c2 — 2 ab — 26c + 2co)
[v (® + y + z)2 = x 2 + y2 + z2 + 2xy + 2yz + 2xz]
= 2o2 + 262 + 2c2 + 4ca
(a + b + c)2 + (a —b + c)2 = 2 a2 + 2b2 + 2c2 + 4ca
Q2.2: (a + b + c)2 — (a —b + c)2
Ans: We have,
(a + b + c)2 — (a —b + c)2 = (a2 + b2 + C2 + 2ab + 26c + 2co)
- (a2 + ( — 6)2 + c2 - 2a6 - 26c + 2ca)
['.• (x + y + z)2 = x 2 + y 2 + z2 + 2 xy + 2 yz + 2 xz]
= a2 + ti2 + c2 + 2ab + 26c + 2ca — a2 — 62 — c2 + 2a6 + 26c — 2co)
= 4o6 + 46c
(a + 6 + c)2 — (a — 6 + c)2 = 4a6 + 46c
Q2.3: (a + b + c)2 + (a + 6 — c)2 + (a + b — c)2
Ans: We have.
(a + b + c)2 + (a + b — c)2 + (a + b — c)2 = a2 + b2 + <? + 2ab + 26c + 2ca
+ (a2 + 62 + (z)2 — 26c — 2o6 + 2ca) + (a2 + 62 + c2 — 2ca — 26c + 2ab)
[•.• (x + y + z)2 = x 2 + y 2 + z2 + 2 xy + 2 yz + 2 xz]
— 3 a2 + 362 + 3c2 + 2a6 + 26c + 2ca — 26c — 2a6 — 2ca — 26c + 2ab
= 3®2 + 3 y 2 + 3 z 2 + 2ab — 26c + 2ca
(a+ b + c)2 + (a + b — c)2 + (a — b + c)2 = 3a2 + 362 + 3c2 + 2 ab — 26c + 2ca
(a + 6 + c)2 + (a + 6 — c)2 + (a — 6 + c)2 = 3(a2 + 62 + c2) + 2(ab — bc + ca)
Q2.4: (2x + p — c)2 — (2x — p + c)2
Arts: We have,
(2® + p - c)2 — (2x - p + c)2 = [2a;2 + p2 + ( - c ) 2 + 2(2®)p + 2 p (-c ) + 2(2®)(— c)]
- [4a;2 + ( - p ) 2 + c2 + 2(2®)(—p) + 2 c (-p ) + 2(2®)c]
[y (® + p + z )2 = ®2 + p2 + z2 + 2xy + 2yz + 2xz]
(2x + p — c)2 — (2a; — p + c)2 = [4®2 + p2 + c2 + 4®p — 2pc — 4®c]
- [4®2 + p2 + c2 — 4®p — 2pc + 4®c]
Opening the bracket,
(2x + p — c)2 — (2x — p + c)2 = 4a;2 + p2 + c2 + 4®p — 2pc — 4c® — 4®2 — p2 — c2 + 4®p
+ 2pc — 4c®]
(2® + p — c)2 — (2® — p + c)2 = 8®p — 8®c
=8 ®(p — c)
Hence, (2® + p — c)2 — (2® — p + c)2 = 8®(p — c)
Q2.5:(x2 + y 2 + ( - z ) 2) - (®2 - y2 + z2) 2
Ans: We have,
(®2 + p 2 + ( - z ) 2) 2 - ( ® 2( - p ) 2 + z 2) 2
[®4 + p4 + ( - z ) 4 + 2x2y 2 + 2 y2( - z ) 2 + 2®2( - z ) 2]
- [®4 + ( - p ) 4 + z4 - 2®2p2 - 2p2z2 + 2®2z 2]
['.• (® + p + z )2 = ®2 + p2 + z2 + 2®p + 2 pz + 2®z]
Taking the negative sign inside,
[®4 + p4 + ( - z ) 4 + 2®2p2 + 2 p2( - z ) 2 + 2®2( - z) 2]
- [®4 + ( - p ) 4 + z4 - 2®2p2 - 2p2z2 + 2®2z 2]
= 4®2p2-4 z 2®2
Hence, (®2 + p2 + ( — z )2) 2 — (®2( —p)2 + z2)2 = 4®2p2-4 z 2®2
Q3: lfa+b+c=0 and a2 + 62 + c2 = 16, find the value o f ab+bc+ca:
Ans: We know that,
[y (a + 6 + c)2 = a2 + 62 + c2 + 2a6 + 26c + 2ca]
(0)2=16+2(ab+bc+ca)2(ab+bc+ca)=-16 ab+bc+ca=-8
Hence, value of required express ab+bc+ca =-8
Q4: If a2 + br2 + c2 = 16 and ab+bc+ca=10, find the value ofa+bn:?
Ans: We know that,
(a + 6 + c)2 = a2 + 62 + c2 + 2(a6 + 6c + ca)(® + p + z )2 = 16 + 2(10)(® + p + z )2 = 36 (® + p + z) = i/36 (® + p + z) = ±6
Hence, value of required expression I; (a + 6 + c) = ±8
Q5: lfa+b+c=9 and ab+bc+ca=23, find value o f a2 + 62 + c2
Ans:\Ne know that,
(a + 6 + c)2 = a2 + 62 + c2 + 2(a6 + be + ca)92 = a2 + 62 + c2 + 2(23)81 = a2 + 62 + c2 + 46 a2 + 62 + c2 = 81 - 46 a2 + 62 + c2 = 35
Hence, value of required expression a2 + b2 + c2 = 35
Q6: Find the value of the equation: Ax2 + y2 + 25 z2 + 4 xy — 10 yz — 20 zx when x=4,y=3^=2
Ans: 4x2 + y2 + 25 z2 + 4xy — 10 yz — 20 zx
(2a:)2 + y2 + ( - 5 z)2 + 2(2 x)(y) + 2(y)(— 5z) + 2(—5z)(2x)(2x + y — 5 z)2 (2(4) + 3 — 5(2))2 (8 + 3 - 10)2 ( l ) 2 1
Hence value of the equation is equals to 1
Q7: Simplify each of the following expressions:
Q7.1:(x + y + z)2 + (x + f + f ) 2 - ( f + f + f ) 2
Ans: Expanding, we get
= [a;2 + y2+ z 2 + 2 xy + 2 yz + 2za;] + [a:2 + ^ + T + 2x| + 2x + f ]
[v (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz]
= x2 + y2 + z2 + 2xy + 2yz + 2zx + a;2 + ^ ^ + 2a:| + ^ ^ - £_ HI _ ££
6 4
Rearranging coefficients,
8a;2—x2 . 36y2+9y2—4y2 144z2+16z2-9z2 . 6xy+3xy-xy 13 yz 29 xz4 ' 36 + 144 "l" 3 + 6 + 12
_ 7if_ , 41y 151z2 , 8xy“ 4 + 36 + 144 + 3
132/3 29zx6 12
(a, + „ + 2)» + (, + | + i ) . _ ( f + I + z). = ^ + a . + ^ + ^ + lfc + ^Q 7.2: (x + y — 2 z)2 — x 2 — y2 — 3 z 2 + 4 xy
Ans: (x + y — 2 z)2 — x 2 —y2 — 3 z2 + 4a:y
= [a:2 + y2 + 4z 2 + 2xy + 2y(-2z) + 2a(-2c)] - x 2 - y 2 - 3z2 + 4xy
— z 2 + 6 xy — 4 yz — 4zx
(x + y — 2 z)2 — x 2 — y2 — 3 z2 + 4 xy = z2 + 6xy — 4 yz — 4 zx
Q 7.3: [x2 - x + l}2 - [x2 + x + l}2
Ans: [x2 — x + l ]2 — [x2 + x + l ]2
= (a:2) 2 + ( -a : )2 + l 2 + 2(a:2) ( -a :) + 2 (-a s )(l) + 2a:2)
— \{x2)2 + a:2 + 1 + 2x 2x + 2a:(l) + 2a:2(l)]
[■.• (x + y + z)2 = x 2 + y2 + z2 + 2 xy + 2 yz + 2xz]
= x4 + y2 + 1 - 2a:3 - 2a: + 2a:2 - a:2 - a:4 - 1 - 2a:3 - 2a: - 2a:2
= -4a:3 - 4a:
= —4a:(a:2 + 1)
Hence simplified equation =[x2 — x + l ]2 — [a:2 + x + l ]2 = — 4a:(a:2 + 1)
RD Sharm a Solutions Class 9 A lgebraic Identities Ex 4.3
RD Sharma Solutions Class 9 Chapter 4 Ex 4.3 Q1. Find the cube o f each of the following binomial expression
w ( 7 + ! )
(b) ( f - J r )(c) ( 2 x + ■§)
< « 0 ( 4 - £ )
S o l:
( a ) ( i + f ) ) 3
Given, ( I + f ) ) 3
The above equation is in the fo rm o f (a + b )3 = a3 + b3 + 3ab(a + b)
We know tha t, a = — , b = -|x 3
By using (a + b)3 fo rm u la
( i + f ) ) 3= ( i ) 3+ ( f ) 3+3 ( i ) ( f ) ( i + ! )
= ^ + £ + 3 * M ( M )
= ^ + £ + f ( i r + f )
+ ( ! * i ) +'.T .T'27
' x 3 '
= j _ + £. + jl + yL)x 3 27 X2 3x >
Hence,
G + l w - i + t + i + £ )
( b ) ( ( f - J r ) ) 3
Given, ( ( | - ^ ) ) 3
The above equation is in th e fo rm o f (a - b )3 = a3 - b3 - 3ab(a - b)
We know tha t, a = | - , b = -4-
By using (a - b)3 fo rm u la
( d - ^ ) ) 3= ( | ) 3- ( ^ ) 3- 3 ( l ) ( ^ ) ( l - | = 2 Z . - J . - 3 * 2 * A r l - A t
x3 X 6 X x2 ' x X 2
27 _ _8_ _ 1 8 / 3 _ _2_\x3 x6 x3 ' x x2
27 _ ± - ( I® * 2 1 + f i i * AX 3 X 6 ' X 3 ® ' ' X 3 X 2
27 _ _ 8_ _ 54 + 36
27 _ _8_ _ 54 36Hence, ( ( f - J r ) ) 3
(c) (2a; + f - ) 3
Given, (2a; + | ) 3
The above equation is in th e fo rm o f (a + b )3 = a3 + b3 + 3ab(a + b)q
We know tha t, a = 2 x , b = —
By using (a + b)3 fo rm u la
= 8x 3 + ^ - + ^ ( 2x + | )
= 8x 3 + 4 - + ^ ( 2 x + | )x 3 X ' X /
= 8x3 + ^ +X 3
( 1 8* 2 x ) + ( 1 8 * | )
= 8x3 + ^ + 36x
Hence,
The cube o f (2a; + - | ) 3 = 8x3 + ^ + 36x
( d ) ( 4 - ^ ) 3
Given, ( 4 - ^ ) 3
The above equation is in th e fo rm o f (a - b )3 = a3 - b3 - 3ab(a - b)
We know th a t ,a = 4 , b = ^
By using (a - b)3 fo rm u la
( 4 - ^ ) 3 = 4 3 - ( ^ ) 3 - 3 ( 4 ) ( - L ) ( 4 - 3 L )
- ^ “ 27x3 ~ 3 x ( 4 “ 3 ^ )
= 64-----------— (4 — —)W 27a;3 * ^ 3 * '
= 64 - 1 - (A * 41 + (A. * J - l27a:3 3a: 3a: 3a:'
= 6 4 ------1___ — + ( —
H e n ce ,
T h e c u b e o f(4 --L )3 = 6 4 - ^ - ^ + ( £ )
Q2. Simplify each of the following
(a) (x + 3) 3 + ( x - 3) 3
( b ) ( f + ! ) 3 - ( f - ! ) 3
(c) (* + f ) 3 + (*“ f ) 3(d) (2x - 5y) 3 - (2x + 5y)3
S o l:
(a) (x + 3 )3 + ( x - 3 )3
The above equation is in th e fo rm o f a3 + b3 = (a + b )(a2 + b2 - ab)
We know tha t, a = (x + 3 ) , b = (x - 3)
By using (a3 + b3) fo rm u la
= (x + 3 + x - 3 )[ (x + 3 )3 + (x - 3 )3 - (x + 3)(x - 3)]
= 2x[(x2 + 3 2 + 2 *x*3 ) + (x2 + 3 2 - 2 *x*3 ) - (x2 - 32)]
= 2x[(x2 + 9 + 6x) + (x2 + 9 - 6x) - x 2 + 9]
= 2x[(x2 + 9 + 6x + x2 - 9 - 6x - x 2 + 9)]
= 2x ( x2 + 27)
= 2x3 + 54x
Hence, the resu lt o f (x + 3 )3 + ( x - 3 )3 is 2x3 + 54x
(b)(f + f ) 3- ( f - | ) 3The above equation is in th e fo rm o f a3 - b3 = (a - b )(a2 + b2 + ab)
We know tha t, a = ( | + | ) 3 ,b = ( f — f ) 3
By using (a3 - b3) fo rm u la
= K(f + -f )3 - ( ( f - f )3 )][((# + f )3)2 ((f - f )3)2 - ( (f + f )3 ) ( ( f - f )3)( f + f - f + ! ) [ ( ( l )2 + ( f ) 2 + (3 r)) + ( ( f ) 2 + ( ! ) 2- ( ^ ) ) + ( ( f )2- ( f ) 2)]2 x y x
yLu* + y~ +3 LV 4 T 9 T
Zn\\22 x y \ _ i_ ( £ _ i y _ _ Zxy \ i V_
f 9 6 ' 4 9
_ r s 2 ■ V 2 ■ . x * . y ^ _ _ 2 x y _ y 2 n
“ 3 U “h 9 “h 6 “, _ 4 “h 9 6 “l~ 4 9 J
- ^ r £ l \ y2 \ x2 \ * i i- 3 L 4 9 4 ~■ 4 J
- ^ [ 3 * 1 i t . ]3 L 4 “l" 9 J
_ s 2y , 2y3 2 27
Hence, the resu lt o f ( f + | ) 3 - ( | - | ) 3 = ^ + ^ -
(c) ( * + l ) 3 + ( * - l ) 3The above equation is in th e fo rm o f a3 + b3 = (a + b )(a2 + b2 - ab)
We know tha t, a = (a: + - | ) 3 , b = (x- 1 -)3
By using (a3 + b3) fo rm u la
= ( * + ! + * - ! ) [ ( * + 1 ) 2 + ( * - i ) 2- ( ( * + ! ) ( * - ! ) ) ]
= (2 a :)[(a ;2 + £ + f ) + ( * 2 + H * 2 - £ )
= ^ [ ( z 2 + 4 r + f + x 2 + + ± )
= (2x) [ ^ + ± + ± + ± )
= {2x)[{x* + ^ )
= 2x3 + f
Hence, the resu lt o f (x + | - ) 3 + (x - | - ) 3 = (2a:) [(a:2 + ^ | )
(d) (2x - 5y)3 - (2x + 5y)3
Given, (2x - 5y)3 - (2x + 5y)3
The above equation is in th e fo rm o f a3 - b3 = (a - b )(a2 + b2 + ab)
We know tha t, a = (2x - 5 y ) , b = (2x + 5y)
By using (a3 - b3) fo rm u la
= (2x - 5y - 2x - 5y)[(2x - 5y)2 + (2x + 5y)2 + ((2x - 5y) * (2x + 5y» ]
= (-10y)[(4x2 + 25y2 - 20xy) + (4x2 + 25y2 + 20xy) + 4x2 - 25y2]
= (-10y)[ 4x2 + 25y2 - 20xy + 4 x 2 + 25y2 + 20xy + 4x2 - 25y2]
= (-10y)[4x2 + 4x2 + 4 x2 + 25y2]
= (-10y)[12x2 + 25y2}
= -120x2y - 250y3
Hence, the resu lt o f (2x - 5y)3 - (2x + 5y)3 = -120x2y - 250y3
Q3. If a + b = 10 and ab = 21, Find the value of a3 + b3
S o l:
Given,
a + b = 10, ab = 21
w e know tha t, (a + b )3 = a3 + b3 + 3ab(a + b) ------- 1
s u b s titu te a + b = 1 0 , ab = 21 in eq 1
=> (10 )3 = a3 + b3 + 3(21 )(1 0)
=> 1000 = a3 + b3 + 630
=> 1000 - 630 = a3 + b3
=> 370 = a3 + b3
Hence, the value o f a 3 + b3 = 370
Q4. If a - b = 4 and ab = 21, Find the value of a3 - b3
S o l:
Given,
a - b = 4 , ab = 21
w e know tha t, (a - b )3 = a3 - b3 - 3ab(a - b) ------- 1
s u b s titu te a - b = 4 , ab = 21 in eq 1
=> (4 )3 = a3 - b3 - 3(21 )(4)
=> 64 = a3 - b3 - 252
=> 64 + 252 = a3 - b3
= > 3 1 6 = a3 - b3
Hence, the value o f a 3 - b3 = 316
Q5. If (a; + = 5 , Find the value o f x 3 + ^
S o l:
Given, (x + j ) = 5
We know tha t, (a + b )3 = a3 + b3 + 3ab(a + b)
S ubstitu te (x + ^ -) = 5 in eq 1
(x + i ) 3 = x 3 + ± + 3 (x * i ) ( x + i )
53 = x3 + ^ + 3(x * I ) ( x + | )
125 = x3 + J3 + 3 ( x + i )
— 1
125 = x3 + ^ + 3 (5 )
125 = x 3 + ^ + 1 5X s
1 2 5 - 15 = x3 + \X A
x 3 + 4 r = 110®3
hence, th e resu lt is x 3 + -V = 1 1 0X s
Q6 . If (x - = 7 , Find the value of x3-
S o l:
Given, If (x- ^ ) = 7
We know tha t, (a - b )3 = a3 - b3 - 3ab(a - b) ------- 1
S ubstitu te (a :- = 7 in eq 1
73 = x3 - ^ a - 3 ( x - | )
343 = x3 - ^ - ( 3 * 7 )
343 = x 3 - - 21X3
343 + 21 = x3 -r-X 3
x 3 - 4 = 3 6 4X 3
hence, th e resu lt is x 3 - = 364X 3
Q7. If ( x - = 5 , Find the value o f a;3- ^
S o l:
Given, If ( x - j ) = 5
We know tha t, (a - b )3 = a3 - b3 - 3ab(a - b) ------- 1
S ubstitu te ( x - - ) = 5 in eq 1
( * - ¥ ) 3 = x3 - : ? - 3 ( x * ¥ ) ( x - f )
53 = x3 - ^ - 3 ( x - i )
1 2 5 = x 3 - ^ - ( 3 * 5 )
125 = x 3 - \ - 15Xs
125 + 15 = x3 - -VX 6
x 3 - = 140X 3
hence, th e resu lt is x 3 - -V = 140X 3
Q8 . If (x2 + = 5 1 , Find the value of x3- ^
S o l:
Given, (x2 + ± ) = 51
We know th a t , ( x - y )2 = x2 + y2 - 2xy ------- 1
S ubstitu te (a:2 + ^ - ) = 51 in eq 1
( x - ^ ) 2 = x 2 + ^ - 2 * x * i
(x - ¥ ) 2 = x 2 + ^ - 2
( x - i )2 = 5 1 - 2
(x - ^ ) 2 = 4 9
( x - I ) = V 3 9
( * - £ > = * 7
We need to find a:3-
So, a3 - b3 = (a - b )(a2 + b2 + ab)
a;3“ ^ = (x _ 7 ) ( xA{2>+ ^ + (x * ¥ )
We know th a t ,
( x - | ) = 7 and (a:2 + = 51
* 3- £ = 7 (5 1 + 1 )
* 3— ^ = 7 ( 5 2 )
a:3- ^ = 364
Hence, the value o f x3- = 364
Q9. If (a;2 + = 9 8 , Find the value o f a;3 + ^
S o l:
Given, (a;2 + ^ ) = 98
We know th a t , ( x + y )2 = x2 + y2 + 2xy ------- 1
S ubstitu te (a;2 + p - ) = 98 in eq 1
(x + - ) 2 = x2 + - V + 2 * x * -
(X + - ) 2 =X 2 + + 2
(x + ^ ) 2 = 9 8 + 2
(x + i ) 2 = 1 0 0
( x + i ) = v / l0 0
( x + I ) =±10
We need to find a;3 + -V
So, a3 + b3 = (a + b )(a2 + b2 - ab)
* 3 + ^ - = ( x + | ) ( x 2 + ^ - - ( x * ^ )
We know that,
( x + ^ ) = 10 and (a;2 + ^ j ) = 98
a:3 + ^ - = 1 0 ( 9 8 - 1 )
a:3 + ^ = 1 0 ( 9 7 )
a;3 + -V = 970
Hence, the value o f a:3 + = 970ar
Q10. If 2x + 3y = 13 and xy = 6, Find the value of 8X3 + 2 7 ^
S o l:
G ive n , 2x + 3y = 1 3 , xy = 6
We know that,
(2x + 3y)3 = 132
=> 8x3 + 27y3 + 3 (2x)(3y)(2x + 3y) = 2197
=> 8x3 + 27y3 + 18xy(2x + 3y) = 2197
S ubstitu te 2x + 3y = 13, xy = 6
=> 8x3 + 27y3 + 18 (6 )(13) = 2197
=> 8x3 + 27y3 + 1404 = 2197
=> 8x3 + 27y3 = 2 1 9 7 - 1404
=> 8x3 + 27y3 = 793
Hence, the value o f 8x3 + Hy3 = 793
Q11. If 3x — 2y =11 and xy= 12, Find the value of 27x3 - 8 y 3
S o l:
Given, 3x - 2y = 11 , x y = 12
We know th a t (a - b)3 = a3 - b3 - 3ab(a + b)
(3x - 2y)3 = 113
= > 27x3 - 8y3 - ( 1 8 * 1 2 * 1 1 ) = 1331
=> 27x3 - 8y3 - 2376 =1331
=> 27x3 - 8y3 = 1331 + 2 3 7 6
=> 27x3 - 8y3 = 3707
Hence, the value o f 27x3 - 8y3 = 3707
Q12. If as4 + ( ^ j ) = 119, Find the value o f z 3- ( ^ )
Sol:
G iv e n ,*4 + ( ^ j ) = 119 ------- 1
We know th a t (x + y )2 = x2 + y2 + 2xy
S ubstitu te * 4 + ( ^ - ) = 119 in eq 1
( * 2 + ( ^ ) ) 2 = x4 + ( £ ) + P**2* j l )
= x4 + ( ^ ) + 2
= 119 + 2
= 121
( * 2 + ( ^ ) ) 2 =121
x 2 + ( ^ ) = ^ 1 2 1
a:2 + ( J _ ) = +11
Now, find (x - -i-)
We know th a t (x - y )2 = x2 + y2 - 2xy
(x - ^ ) 2 = x2+ ^ - ( 2 * x4
= x2 + A “ 2X 1
= 11-2
= 9
( * - £ ) = > /§
= ±3
We need to find * 3- ( ^ - )
We know tha t, a3 - b3 = (a - b )(a2 + b2 - ab)
* 3- ( ^ ) = ( x - 7 X * 2 + ( ^ ) + x 4
Here, x2 + (■^■) = 11 and (x - = 3
z 3- ( ^ ) = 3 (1 1 + 1 )
= 3(12)
= 36
Hence, the value o f x3- (■^■) = 36
Q13. Evaluate each of the following
(a) (103)3
(b ) (98)3
(c) (9.9) 3
(d) (10.4)a
(e) (598)3
(f)(9 9 ) 3
S o l:
Given,
(a) (103 )3
w e know th a t (a + b )3 = a3 + b3 + 3ab(a + b)
=> (103)3 can be w ritte n as (100 + 3 )3
H e re , a = 100 and b = 3
(103)3 = (100 + 3 )3
= (100 )3 + (3 )3 + 3 (100 )(3 )(100 + 3)
= 1 0 0 0 0 0 0 + 2 7 + (900*103)
= 1000000 + 27 + 92700
= 1092727
The va lue o f (10 3 )3 = 1092727
(b) (98 )3
w e know th a t (a - b )3 = a3 - b3 - 3ab(a - b)
=> (98 )3 can be w ritte n as (100 - 2 )3
H e re , a = 100 and b = 2
(98 )3 = ( 1 0 0 - 2 ) 3
= (100 )3 - (2 )3 - 3 (100 )(2 )(100 - 2)
= 1 0 0 0 0 0 0 - 8 - ( 6 0 0 * 1 0 2 )
= 1 0 0 0 0 0 0 - 8 - 58800
= 941192
The va lue o f (98 )3 = 941192
(c) (9.9)3
w e know th a t (a - b )3 = a3 - b3 - 3ab(a - b)
=> (9 .9)3 can be w ritte n as (10 - 0.1 ) 3
H e re , a = 10 and b = 0.1
(9 .9)3 = ( 1 0 - 0 .1 )3
= (10 )3 - (0.1 ) 3 - 3 (10)(0.1 )(1 0 - 0.1)
= 1000 - 0.001 - (3*9.9)
= 1 0 0 0 -0 .0 0 1 - 2 9 .7
= 1 0 0 0 -2 9 .7 0 1
= 970.299
The va lue o f (9 .9 )3 = 970.299
(d) (10 .4 )3
w e know th a t (a + b )3 = a3 + b3 + 3ab(a + b)
=> (10 .4 )3 can be w ritte n as (10 + 0 .4 )3
H e re , a = 10 and b = 0.4
(10 .4 )3 = (10 + 0 .4 )3
= (10 )3 + (0 .4 )3 + 3 (10)(0 .4 )(10 + 0.4)
= 1 0 0 0 + 0 .0 6 4 + (12*10.4)
= 1000 + 0 .064 + 124.8
= 1000 + 124.864
= 1124.864
The va lue o f (10 .4 )3 = 1124.864
(e) (598)3
w e know th a t (a - b )3 = a3 - b3 - 3ab(a - b)
=> (598)3 can be w ritte n as (600 - 2 )3
H e re , a = 600 and b = 2
(598)3 = (600 - 2) 3
= (600)3 - (2 )3 - 3 (600)(2 )(600 - 2)
= 216000000 - 8 - (3600*598)
= 216000000 - 8 - 2152800
= 216000000 - 2152808
= 213847192
The va lue o f (598 )3 = 213847192
( f) (9 9 )3
w e know th a t (a - b )3 = a3 - b3 - 3ab(a - b)
=> (99 )3 can be w ritte n as ( 1 0 0 - 1 ) 3
H e re , a = 100 and b = 1
(99 )3 = ( 1 0 0 - I ) 3
= (100 )3 - ( I ) 3 - 3 (1 00 )(1 )(100 - 1)
= 1000000 - 1 - (300*99)
= 1 0 0 0 0 0 0 - 1 -2 9 7 0 0
= 1 0 0 0 0 0 0 -2 9 7 0 1
= 970299
The va lue o f (99 )3 = 970299
Q14. Evaluate each of the following
(a) 1113 - 893
(b) 463 + 343
(c) 1043 + 963
(d) 933 - 1073
Sol:
Given,
(a) 1113 - 893
the above equation can be w ritte n as (100 + 11 ) 3 - (100 - 11 ) 3
w e know tha t, (a + b )3 - (a - b )3 = 2 [b3 + 3ab2]
here, a= 100 b = 11
(100 + 11 ) 3 - (100 - 11 )3 = 2 [113 + 3 (100 )2(1 1)]
= 2 [1331 + 3 3 0 0 0 0 ]
= 2(331331]
= 662662
The va lue o f 1113 - 89 3 = 662662
(b) 4 6 3 + 343
the above equation can be w ritte n as (40 + 6 )3 + (40 - 6 )3
w e know tha t, (a + b )3 + (a - b )3 = 2 [a3 + 3ab2]
here, a= 4 0 , b = 4
(40 + 6 )3 + (40 - 6 )3 = 2 [403 + 3 (6 )2(40)]
= 2(64000 + 4320]
= 2(68320]
= 1366340
The va lue o f 46 3 + 343 = 1366340
(c) 1043 + 963
the above equation can be w ritte n as (100 + 4 )3 + ( 1 0 0 - 4 )3
w e know tha t, (a + b )3 + (a - b )3 = 2 [a3 + 3ab2]
here, a= 100 b = 4
(100 + 4 )3 - (100 - 4 )3 = 2 [1003 + 3 (4 )2(100)]
= 2(1000000 + 4800]
= 2(1004800]
= 2009600
The va lue o f 1043 + 96 3 = 2009600
(a) 933 - 107 3
the above equation can be w ritte n as (100 - 7 )3 - (100 + 7 )3
w e know tha t, (a - b )3 - (a + b )3 = -2 [b3 + 3ba2]
here, a= 93, b = 107
(100 - 7)3 - (100 + 7 )3 = -2[73 + 3 (100 )2(7)]
= -2(343 + 210000]
= -2(210343]
= -420686
The va lue o f 933 - 1073 = -420686
Q15. I f x + ^ = 3, calculate x 2 + + \ , x i + \X x a X4
S o l:
Given, x + — = 3X
We know th a t (x + y )2 = x 2 + y2 + 2xy
( x + ^ ) 2 = x 2 + ^ + ( 2 * * * 1 )
32 = x 2 + + 2X2
9 - 2 = x 2 + 4X2
x2+ 4 = 7
squaring on bo th s ides
( x2 + ^ ) 2 = 72
x4 + A . + 2 * x2 * ^ = 4 9a;4 x 1
X4 + J j . + 2 = 49x i
x4 + A - = 4 9 - 2X 4
x4 + 4 - = 4 7
a g a in , cub ing on bo th s ides
(x + ^ ) 3 = 33
x 3 + + 3 x *—(x + —) = 27
x 3 + ± + (3 *3 ) = 27
x 3 + J _ + g = 2 7x »
x 3 + \ = 2 7 - 9X A
x3+ \ = 1 8X s
hence, th e va lues are x 2 + = 7, x4 + -V = 47, x 3 + = 1 8X1 x 4 X S
Q16. If x4 + —7 = 194, calculate x 2 + ^ , x 3 + \ , x + ~X 4 X £ X 3 x
S o l:
Given,
x4 + 4 - = 194 --------1X 4
add and su b tra c t (2 *x 2 * - ^ ) on le ft s ide in above given equationX 1
x4 + —7- + (2 * x 2 * ^ ) - 2 (2 *x 2 * ± ) = 1 9 4
x4 + J_ + (2*a:2 * ^ ) - 2 = 194
(x2 + ~ j )2 “ 2 =194
(x2 + ^ ) 2 = 1 9 4 + 2
(x2+ ^ ) 2 = 196
(x2 + ^ ) = V ^ 9 6
(x2+ Jj ) = 14 -------- 2
A dd and su b tra c t (2 *x* on le ft s ide in eq 2
(x2 + ^ ) + ( 2 * x * i ) - ( 2 * x * | ) = 14
( x + ± ) 2 - 2 = 14
( x + i )2 = 1 4 + 2
( x + ^ ) 2 = 1 6
( x + i ) = 7 1 6
(x+ 7 ) =4 ------- 3
Now, cub ing eq 3 on bo th s ides
( x + i )3 = 4 3
We know tha t, (a + b )3 = a3 + b3 + 3ab(a + b)
x3+ ^ t + 3*x* ^ ( x + ^ ) = 64
x 3 + p r + (3 *4 ) = 64
x 3 + J = = 6 4 - 1 2X 3
x 3 + = 52X A
hence, th e va lues o f (x2 + ■ ■)2 = 196, (x + - j ) = 4 , x 3 + ^ = 5 2
Q17. Find the values o f 27x3 + By3 , if
(a) 3x + 2y = 14 and xy = 8
(b) 3x + 2y = 20 and xy = ^
Sol:
(a) Given, 3x + 2y = 14 and xy = 8
cub ing on both s ides
(3x + 2y)3 = 14 3
We know tha t, (a + b )3 = a3 + b3 + 3ab(a + b)
27x3 + 8y3 + 3 (3x)(2y)(3x + 2y) = 2744
27x3 + 8y3 + 18xy(3x + 2y) = 2744
27x3 + 8y3 + 18(8)(14) = 2744
27x3 + 8y3 + 2016 = 2744
27x3 + 8y3 = 2 7 4 4 - 2 0 1 6
27x3 + 8y3 = 728
Hence, the value o f 27x3 + Sy3 = 728
(b) Given, 3x + 2y = 20 and xy = - y
cub ing on both s ides
(3x + 2y)3 = 203
We know tha t, (a + b )3 = a3 + b3 + 3ab(a + b)
27x3 + 8y3 + 3 (3x)(2y)(3x + 2y) = 8000
27x3 + 8y3 + 18xy(3x + 2y) = 8000
27x3 + 8y3 + 18 ( ^ ) ( 2 0 ) = 8000
27x3 + 8y3 + 560 = 8000
27x3 + 8y3 = 8000 - 560
27x3 + 8y3 = 7440
Hence, the value o f 27x3 + 8y3 = 7440
Q18. Find the value o f 64X3 - 1 25z3 , if 4x - 5z = 16 and xz = 12
Sol:
Given, 64x3 - 125z3
Here, 4 x - 5z = 16 and xz = 12
Cubing 4x - 5z = 16 on bo th s ides
(4x - 5z)3 = 163
We know tha t, (a - b )3 = a3 - b3 - 3ab(a - b)
(4x)3 - (5z)3 - 3 (4x)(5z)(4x - 5z) = 163
64x3 - 125z3 - 60 (xz)(16) = 4096
64x3 - 125z3 - 60(12 )(1 6) = 4096
64x3 - 125z3 - 11520 = 4096
64x3 - 125z3 = 4 096 + 11520
64x3 - 125z3 = 1 5 6 1 6
The va lue o f 64x3 - 125z3 = 15616
Q19. If x - ^ = 3 + 2 \ / 2 , Find the value of x3 - ^
S o l:
Given, x - ^ = 3 + 2 \ / 2
Cubing x - = 3 + 2y/2 on both s ides
We know tha t, (a + b )3 = a3 + b3 + 3ab(a + b)
(x - I ) 3 = (3 + 2 ^ ) 3
x 3 - ± - 3 * x *± (x - | ) = 3 2 + (2-y/2)3+ 3 *3 *2 V ^ ( 3 + 2 i / 2 )
x 3 - 4 - " 3 (3 + 2 v ^ ) = 27 + 16 \ / 2 + 18 v ^ ( 3 + 2-s/2)
x 3 - 4 r = 27 + 16-v/2 + 5 4 + 72 + 9 + 6 -\/2X A
x 3 - ^ = 108 + 76 %/2X s
hence, th e va lue o f x3 - = 108 + 7 6 1/2X *
RD Sharma Solutions Class 9 Algebraic Identities Ex 4.4
RD Sharma Solutions Class 9 Chapter 4 Ex 4.4Q1. Find the following products
(a) (3x + 2y)(9x2 - 6xy + 4y2)
(b) (4x - 5y)(16x2 + 20xy + 25y2)
(c) (7p4 + q)(49p8 - 7p4q + q2)
(d ) ( f +2y)(
T - x y + 4y2)
( 3 5 \ / 9 . 25 , 15 \
(f)(3 + | ) (9 -^ + f )
(g) ( ! + 3 a : ) ( £ + 9 a ! 2-6 )
(h) ( § - 2 * 2)(-^- + 4 x 4- 6 x )
(i) (1 - x)(1 + x + xA{2})
Q)(1 +x)(1 - x + xA{2})
(^(a^-lKa^ + ^ + l)
(I) ( x 2 + l ) ( x 6- x 3 + 1)
Sol:
(a) (3x + 2y)(9x2 - 6xy + 4y2)
Given, (3x + 2y)(9x2 - 6xy + 4y2)
We know that, a3 + b3 = (a + b)(a2 + b2 - ab)
(3x + 2y)(9x2 - 6xy + 4y2) can we written as
=> (3x + 2y)[(3x)2 - (3x)(2y) + (2y)2)]
=> (3x)3 + (2y)3
=> 27x3 + 8y3
Hence, the value o f (3x + 2y)(9x2 - 6xy + 4y2) = 27x3 + 8y3
(b) (4x - 5y)(16x2 + 20xy + 25y2)
Given, (4x - 5y)(16x2 + 20xy + 25y2)
We know that, a3 - b3 = (a - b)(a2 + b2 + ab)
(4x - 5y)(16x2 + 20xy + 25y2) can we written as
=> (4x - 5y)[(4x)2 + (4x)(5y) + (5y)2)]
=> (4x)3 - (5y)3
=> 16x3 - 25y3
Hence, the value o f (4x - 5y)(16x2 + 20xy + 25y2) = 16x3 - 25y3
(c) (7p4 + q)(49p8 - 7p4q + q2)
Given, (7p4 + q)(49p8 - 7p4q + q2)
We know that, a3 + b3 = (a + b)(a2 + b2 - ab)
(7p4 + q)(49p8 - 7p4q + q2) can be written as
=> (7p4 + q)[(7p4)2 - (7p4)(q) + (q)2)]
=> (7p4)3 + (q)3=> 343p12 + q3
Hence, the value o f (7p4 + q)(49p8 - 7p4q + q2) = 343p12 + q3
(d) ( f + 2y)( - xy + 4y2)
Sol:
Given, ( f + 2 y ) ( ^ - x y + 4y2)
We know that, a3 + b3 = (a + b)(a2 + b2 - ab)
( f + 2y)( ^ - xy + 4y2) can be written as
= > ( f + 2 3/) [ ( f ) 2- f ( 2 t /) + (2t/)2]
= > ( f ) 3 + (2 j, )3
=> x + 8y 3
* ) ( * + 7 + S *)
Sol:
G iv e n , ( | - f ) ( A + l | + i | )
We know that, a3 - b3 = (a - b)(a2 + b2 + ab)
/_3__5_\/_9__ |_ 25_ . 15 \' x y )\ x2 ' y 2 ' x y )
Can be written as,
= > ( ! - f ) ( ! ) 2+(f)2+ ( ! ) ( f )
=>(f)3- ( f )3
= > ( § ) - ( ? )
Hence, the value o f ( * - £ ) ( £ + f + § ) = ( f M f )
( f) (3 + | ) ( 9 - f + f )
Sol:
Given,(3 + f ) ( 9 - f + f )
We know that, a3 + b3 = (a + b)(a2 + b2 - ab)
(3 + § -)(9 - + J r ) can be written as,
= > ( 3 + § ) [ ( 3 2) - 3 ( f ) + ( § ) 2]
=>(3)3 + ( | ) 3= > 2 7+ * f
Hence, the value o f (3 + | ) ( 9 - £ + f ) is 27 + ^
( f l ) ( ! + 3 » ) ( ^ + 9 * 2- 6 )
Sol:
Given, ( J + 3a;)( J r + 9a;2- 6 )
We know that, a3 + b3 = (a + b)(a2 + b2 - ab)
(■J + 3a:)(-J- + 9a:2- 6) can be written as,
= > ( ! + 3 a : ) [ ( f ) 2 + (3a:)2- ( | ) ( 3 a : ) ]
= > ( | ) 3 + (3 x )3
= > J r + 9 * 3
Hence, the value o f (-J + 3a;) ( - ^ + 9a:2- 6) is J r + 9a:3
(h ) ( f - 2 a : 2) ( ^ - + 4 a : 4-6 a ;)
Sol:
Given, ( J -2 a ;2) ( J r + 4a:4-6 a :)
We know that, a3 - b3 = (a - b)(a2 + b2 + ab)
(■ § -2a:2)(-J- + 4a:4-6 a :) can be written as,
= > ( ! - 2 * 2) [ ( ! ) 2 + (2a;2)2- ( | ) ( 2 a : 2)]
= > ( f - 2 a ;2) [ ( J r ) + 4 a :4- ( f ) ( 2 a ; 2)]
= > ( f ) 3- (2 a :2) 3
=> ? j - 8 x 6
Hence, ( J - 2 a ;2)(-J- + 4a:4-6 a :) is | j - 8 a : 6
(i) (1 - x)(1 + x + xA{2})
Sol:
Given, (1 - x)(1 + x + xA{2})
We know that, a3 - b3 = (a - b)(a2 + b2 + ab)
(1 - x)(1 + x + xA{2}) can be written as,
=>(1 -x ) [(1 2 + (1)(x)+x2)]
=>(1)3 - ( x ) 3
=> 1 - X3
Hence, the value o f (1 - x)(1 + x + xA{2}) is 1 - x3
( j) (1 +x)(1 - x + xA{2})
Sol:
Given, (1 + x)(1 - x + xA{2})
We know that, a3 + b3 = (a + b)(a2 + b2 - ab)
(1 + x)(1 - x + xA{2}) can be written as,
=>(1+x)[(12 -(1 )(x ) + x2)]
=> (1)3 + (x)3
=> 1 + X3
Hence, the value o f (1 + x)(1 + x - xA{2}) is 1 + x3
(k) (x 2- l) (a ;4 + x 2 + 1)
Sol:
Given, (x 2- 1) (a:4 + x 2 + 1)
We know that, a3 - b3 = (a - b)(a2 + b2 + ab)
(a:2- l) (a :4 + as2 + 1) can be written as,
=>(x2-1 ) [(x 2)2 - 1 2 + (x2)(1)]
=> (x2)3 - 13
=> x6 - 1
Hence, (a;2- l) (a ;4 + a;2 + 1) is x6 - 1
(l) ( x2 + l ) ( x 6- x 3 + 1)
Sol:
Given, ( x2 + l ) ( x 6- x 3 + 1)
We know that, a3 + b3 = (a + b)(a2 + b2 - ab)
(a:2 + l) (a :6-a ;3 + 1) can be written as,
=>(x3 + 1)[(x3)2 - ( x 3)(1) + 12]
=> (x3)3 + 13
=> X9 + 1
Hence, the value o f (a;2 + l) (a ;6-a :3 + 1) is x9 +1
Q2. Find x= 3 and y = -1, Find the values o f each of the following using in identity:
(a) (9x2 - 4x2)(81 y4 + 36x2y2 + 16x4)
/la / 3 5 \ / 9 , 2 5 , 1 5 \
r~\rx v \ ( ^ i v2 xv\(c ) ( 7 + 3 ) ( 49 + 9 21 )
/ / x y x/ a:2 , v2 , xv(d) ( l “ 3 ) ( l6 + T + 2 l)
(e) (| + 5 x ) ( f - 2 5 + 25a:2)
Sol:
(a) (9x2 - 4x2)(81 y4 + 36x2y2 + 16x4)
We know that, a3 - b3 = (a - b)(a2 + b2 + ab)
(9x2 - 4x2)(81 y4 + 36x2y2 + 16X4) can be written as,
=> (9x2 - 4x2)[(9y2)2 + (9)(4)x2y2 + (4x2)2]
=> (9y2)3 - (4x2)3
=> 729y6 - 64x6
Substitute the value x = 3, y = -1 in 729y6 - 64x6 we get,
=> 729y6 - 64x6
=> 729(-1)6 - 64(3)6
=> 729(1) - 64(729)
=> 729 - 46656
=> -45927
Hence, the product value o f (9x2 - 4x2)(81y4 + 36x2y2 + 16x4) = -45927
■ 2t) | 15 \«2 “T xv )xy J
Sol:
Given,
We know that, a3 - b3 = (a - b)(a2 + b2 + ab)
/_3_ 5\/_9__ |_ 25 _|_ 15 \\x y ) y x2 ^ ~ y2 ^ ~ Xy)
Can be written as,
= > ( ! - f ) [ ( f ) 2 + ( f ) 2 + ( ! ) ( ! ) ]
= > ( | ) 3- ( f ) 3
= > ( f ) - ( # ) - - • >
Substitute x = 3 in eq 1
=> 0
Hence, the value ° f ( f - f ) ( J r + | r + i f ) is 0
/„W I , V \( X* , V* XV\(cH t + j ) ( i 8 + y _ n )
Given,
We know that, a3 + b3 = (a + b)(a2 + b2 - ab)
(2L+ V ) ( J + £ - 2 . )
Can be written as,
= > ( f + ! ) [ ( f ) 2 + ( | ) 2- ( f ) ( | ) ]
= > ( f ) 3 + ( | ) 3
=>(^) + ( S ) - - 1Substitute x = 3, y = -1 in eq 1
V 343 > ~ \ 27 >
= > ( -» W J_)V 343 > 27 7
Taking least common multiple, we get
27*27 1*343-> 343*27 27*343
-> 729 3439261 9261
__ 729-343 9261
386 - 9261
Hence, the value o f ( f + f ) ( f f +
/a\ <x v\rx 2 _i » 2 W ( 4 3 ) ( 16 9 21 )
xy_ \ _ 386 21 > ~ 9261
Sol:
Given,
We know that, a3 - b3 = (a - b)(a2 + b2 + ab)
( E - l ) ( £ + £ + 2 )1 4 3 16 T 9 T 21 )
Can be written as,
= > ( f - ! ) [ ( f ) 2 + ( ! ) 2 + ( f ) ( 1 ) ]
= > ( f ) 3- ( ! ) 3
V 64 / ' 27 >
Substitute x = 3, y = -1 in eq 1
V 343 ' v 27 '
= > (I ) + (w)Taking least common multiple, we get
_ > 27*27 , 1*6464*27 ~l~ 27*64
_ 729 , 64" 1728 1728
729+64 -> 1728
_ 7939261
Hence, the value o f + £ + £ ) = ^
(e)(| + 5 x ) ( f - 2 5 + 25a:2)
Sol:
We know that, a3 + b3 = (a + b)(a2 + b2 - ab)
(■| + 5 x )(-p — 25 + 25a;2) can be written as,
= > ( ! + 5x) [ ( ! )2 + (5x )2- ( ! ) ( 5a;)]
= > ( | ) 3 + (5 x )3
=> + 125a:3 ----- 1
Substitute x = 3, in eq 1
=> ™ + 125(3)3
=>2§ + 125 * 27
=>2§ + 3375
Taking least common multiple, we get
__ 125 . 3375*27 27 27*1
_ 125 , 9112527 27
125+91125 -> 27
_ 91250 25
Hence, the value o f ( - | + 5a;) (-p— 25 + 25a;2) is 91230
Q3. If a + b = 10 and ab = 16, find the value o f a2 - ab + b2 and a2 + ab + b2
Sol:
Given, a + b = 10, ab = 16
We know that, (a + b)3 = a3 + b3 + 3ab(a + b)
=> a3 + b3 = (a + b)3 - 3ab(a + b)
=> a3 + b3 = (10)3 - 3(16)(10)
=> a3 + b3 = 1000 - 480
=> a3 + b3 = 520
Substitute, a3 + b3 = 520, a + b = 10 in a3 + b3 = (a + b)(a2 + b2 - ab)
a3 + b3 = (a + b)(a2 + b2 - ab)
520 = 10(a2 + b2 - ab)
^ = ( a 2 + b2 -a b )
=> (a2 + b2 - ab) = 52
Now, we need to find (a2 + b2 + ab)
Add and subtract 2ab in a2 + b2 + ab
=> a2 + b2 + ab - 2ab + 2ab
=> (a + b)2 - ab
Substitute a + b = 10, ab
=> a2 + b2 + ab = 102 - 16
= 1 0 0 -1 6
= 84
Hence, the values of (a2 + b2 - ab) = 52 and (a2 + b2 + ab) = 84
Q4. If a + b = 8 and ab = 6, find the value o f a3 + b3
Sol:
Given, a + b = 8 and ab = 6
We know that, a3 + b3 = (a + b)3 - 3ab(a + b)
=> a3 + b3 = (a + b)3 - 3ab(a + b)
=> a3 + b3 = (8)3 - 3(6)(8)
=> a3 + b3 = 512 - 144
=> a3 + b3 = 368
Hence, the value o f a3 + b3 is 368
Q5. If a - b = 6 and ab = 20, find the value of a3 - b3
Sol:
Given, a - b = 6 and ab = 20
We know that, a3 - b3 = (a - b)3 + 3ab(a - b)
=> a3 - b3 = (a - b)3 + 3ab(a - b)
=> a3 - b3 = (6)3 + 3(20)(6)
=> a3 - b3 = 216 + 360
=> a3 - b3 = 576
Hence, the value o f a3 - b3 is 576
Q6. If x = -2 and y = 1, by using an identity find the value o f the following:
(a) (4y2 - 9x2)(1 6y4 + 36x2y2 + 81 x4)
(b) ( | - f ) ( ^ + T + 1)(c) (by + y )(25 j/2- 7 5 + ^ j r )
Sol:
Given,
(a) (4y2 - 9x2)(1 6y* + 36x2y2 + 81 x4)
We know that, a3 - b3 = (a - b)(a2 + b2 + ab)
(4y2 - 9x2)(1 by4 + 36x2y2 + 81 x4) can be written as,
=> (4y2 - 9x2)[(4x)2 + 4y2*9x2 + (9x2)2)
=> (4y2)3 - (9x2)3
=> 64y6 - 729x6 --------1
Substitute x = -2 and y = 1 in eq 1
=> 64y6 - 729x6
=> 64(1 )6 - 729(-2)6
=> 64 - 729(64)
=> 64(1 - 729)
=> 64(-728)
=> -46592
Hence, the value o f (4y2 - 9x2)(1 by4 + 36x2y2 + 81 x4) is -46592
(b) ( f - f ) ( ^ + T + 1)
here x = -2
We know that, a3 - b3 = (a - b)(a2 + b2 + ab)
( ! - f-)(Jr + T + 1) can be witten as-
= > ( M ) [ ( f ) 2 + ( f ) 2 + ( ! ) ( ! ) ]
= > ( | ) 3- ( f ) 3
= > ( £ ) - ( £ ) — 1
Substitute x = -2 in eq 1
=> -1 +1
=> 0
Hence, the value o f ( | — f )(-p- + y + 1) is 0
( c ) ( 5 y + f ) ( 2 5 y 2- 7 5 + ^ )
Sol:
We know that, a3 + b3 = (a + b)(a2 + b2 - ab)
(5 y + y ) ( 2 5 t / 2- 7 5 + ^y -) can be written as,y y
= > (5 y + f) [(5 y )2 + ( f ) 2 - ( 5 y ) ( f ) ]
=>(5y)3 + ( f ) 3=>125y3 + ( ^ ) ------ 1
y3
Substitute y = 1 ineq 1
=>125(1)3 + ( ^ )
=>125 + 3375
=> 3500
Hence, the value o f (5y + y ) (25s/2— 75 + ) is 3500.
EXERCISE 4.5
Q1. Find the following products:
(a) (3x + 2y + 2z)(9x2 + 4y2 + 4z2 - 6xy - 4yz - 6zx)
(b) (4x - 3y + 2z)(16x2 + 9y2 + 4z2 + 12xy + 6yz - 8zx)
(c) (2a - 3b - 2c)(4a2 + 9b2 + 4c2 + 6ab - 6bc + 4ca)
(d) (3x - 4y + 5z)(9x2 + 16y2 + 25z2 + 12xy - 15zx + 20yz)
Sol:
Given,
(a) (3x + 2y + 2z)(9x2 + 4y2 + 4z2 - 6xy - 4yz - 6zx)
we know that,
x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - zx)
so,
(3x + 2y + 2z)(9x2 + 4y2 + 4z2 - 6xy - 4yz - 6zx) = (3x)3 + (2y)3 + (2z)3 - 3(3x)(2y)(2z)
= 27x3 + 8y3 + 8z3 - 36xyz
Hence, the value o f (3x + 2y + 2z)(9x2 + 4y2 + 4z2 - 6xy - 4yz - 6zx) is 27x3 + Sy3 + 8z3 - 36xyz
(b) (4x - 3y + 2z)(16x2 + 9y2 + 4z2 + 12xy + 6yz - 8zx)
we know that,
x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - zx)
so.
(4x - 3y + 2z)(16x2 + 9y2 + 4z2 + 12xy + 6yz - 8zx) = (4x)3 + (-3y)3 + (2z)3 -3(4x)(-3y)(2z)
= 64x3 - 27y3+ 8z3 + 72xyz
Hence, the value o f (4x - 3y + 2z)(16x2 + 9y2 + 4z2 + 12xy + 6yz - 8zx) is 64x3 - 27y3+ 8z3 + 72xyz
(c) (2a - 3b - 2c)(4a2 + 9b2 + 4c2 + 6ab - 6bc + 4ca)
we know that,
x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - zx)
so,
(2a - 3b - 2c)(4a2 + 9b2 + 4c2 + 6ab - 6bc + 4ca) = (2a)3 + (-3b)3 + (-2c)3 - 3(2a)(-3b)(-2c)
= 8a3 - 27b3 - 8c3 - 36abc
Hence, the value o f (c) (2a - 3b - 2c)(4a2 + 9b2 + 4c2 + 6ab - 6bc + 4ca) is 8a3 - 27b3 - 8c3 - 36abc
(d) (3x - 4y + 5z)(9x2 + 16y2 + 25z2 + 12xy - 15zx + 20yz)
we know that,
x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - zx)
so,
(3x - 4y + 5z)(9x2 + 16y2 + 25z2 + 12xy - 15zx + 20yz) = (3x)3 + (-4y)3 + (5z)3 -3(3x)(-4y)(5z)
= 27x3 - 64y3 + 125z3 + 180xyz
Hence, the value o f (3x - 4y + 5z)(9x2 + 16y2 + 25z2 + 12xy - 15zx + 20yz) is 27x3 - 64y3 + 125z3 + 180xyz
Q2. I fx + y + z = 8 and xy + yz + zx = 20, Find the value of x3 + y3 + z3 - 3xyz
Sol:given, x + y + z = 8 and xy + yz + zx = 20
We know that,
(x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx)
(x + y + z)2 = x2 + y2 + z2 + 2(20)
(x + y + z)2 = x2 + y2 + z2 + 40
82 = x2 + y2 + z2 + 40
64 - 40 = x2 + y2 + z2
x2 + y2 + z2 = 24
we know that,
x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - zx)
x3 + y3 + z3 - 3xyz = (x + y + z)[(x2 + y2 + z2) - (xy + yz + zx)]
here, x + y + z = 8, xy + yz + zx = 20, x2 + y2 + z2 = 24
x3 + y3 + z3 - 3xyz = 8[(24 - 20)]
= 8 * 4
= 32
Hence, the value o f x3 + y3 + z3 - 3xyz is 32
Q3. Ifa + b + c = 9 and ab + be + ca = 26, Find the value o f a3 + b3 + c3 - 3abc
Sol:
Given, a + b + c = 9 and ab + be + ca = 26
We know that,
(a + b + c)2 = a2 + b2 + c2 + 2(ab + be + ca)
(a + b + c)2 = a2 + b2 + c2 + 2(26)
(a + b + c)2 = a2 + b2 + c2 + 52
92 = a2 + b2 + c2 + 52
81 - 52 = a2 + b2 + c2
a2 + b2 + c2 = 29
we know that.
a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - be - ca)
a3 + b3 + c3 - 3abc = (a + b + c)[(a2 + b2 + c2) - (ab + be + ca)]
here, a + b + c = 9, ab + be + ca = 26, a2 + b2 + c2 = 29
a3 + b3 + c3 - 3abc = 9[(29 - 26)]
= 9 * 3
= 27
Hence, the value o f a3 + b3 + c3 - 3abc is 27
Q4. Ifa + b + c = 9 and a2 + b2 + c2 = 35, Find the value o f a3 + b3 + c3 - 3abc
Sol:
Given, a + b + c = 9 and a2 + b2 + c2 = 35
We know that,
(a + b + c)2 = a2 + b2 + c2 + 2(ab + be + ca)
92 = 35 + 2(ab + be + ca)
81 = 35 + 2(ab + be + ca)
81 - 35 = 2(ab + be + ca)
■y = ab + be + ca
ab + be + ca = 23
we know that,
a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - be - ca)
a3 + b3 + c3 - 3abc = (a + b + c)[(a2 + b2 + c2) - (ab + be + ca)]
here, a + b + c = 9, ab + be + ca = 23, a2 + b2 + c2 = 35
a3 + b3 + c3 - 3abc = 9[(35 - 23)]
= 9 *1 2
= 108
Hence, the value o f a3 + b3 + c3 - 3abc is 108
Q5. Evaluate:
(a) 253 - 753 + 503
(b) 483 - 303 - 183
(c ) ( ^ ) 3 + ( | ) 3- ( ! ) 3
(d) (0.2)3 - (0.3)3 + (0.1 )3
Sol:
Given,
(a) 253 - 753 + 503
we know that,
a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - be - ca)
here, a = 25, b = -75, c = 50
a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - be - ca)
a3 + b3 + c3 = (a + b + c)(a2 + b2 + c2 - ab - be - ca) + 3abc
a3 + b3 + c3 = (25 - 75 + 50)(a2 + b2 + c2 - ab - be - ca) + 3abc
a3 + b3 + c3 = (0)(a2 + b2 + c2 - ab - be - ca) + 3abc
a3 + b3 + c3 = 3abc
253 + (-75)3 + 503 = 3abc
= 3(25)(-75)(50)
= -281250
Hence, the value 253 + (-75)3 + 503 = -281250
(b) 483 - 303 - 183
we know that,
a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - be - ca)
here, a = 48, b = -30, c = -18
a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - be - ca)
a3 + b3 + c3 = (a + b + c)(a2 + b2 + c2 - ab - be - ca) + 3abc
a3 + b3 + c3 = (48 - 30 - 18)(a2 + b2 + c2 - ab - be - ca) + 3abc
a3 + b3 + c3 = (0)(a2 + b2 + c2 - ab - be - ca) + 3abc
a3 + b3 + c3 = 3abc
483 + (-30)3 + (-18)3 = 3abc
= 3(48)(-30)(-18)
= 77760
Hence, the value 483 + (-30)3 + (-18)3 = 77760
( c ) ( i ) 3 + ( | ) 3- ( ! ) 3
we know that,
a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - be - ca)
here, a= | , b = | , c = ^
a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - be - ca)
a3 + b3 + c3 = (a + b + c)(a2 + b2 + c2 - ab - be - ca) + 3abc
a3 + b3 + c3 = (- j +-| - - |)( a2 + b2 + c2 - ab - be - ca) + 3abc
by using least common multiple
a3 + b3 + c3 = ( ^ | + f j§ ) ( a2 + b2 + c2 - ab - be - ca) + 3abc
a3 + b3 + c3 = + j 2~ - j f ) ( a2 + b2 + c2 - ab - be - ca) + 3abc
a3 + b3 + c3 = 0( a2 + b2 + c2 - ab - be - ca) + 3abc
a3 + b3 + c3 = 3abc
( | ) 3 + ( l ) 3 - ( ^ ) 3 = 3 * f 4 *- 56
Hence, the value o f ( | )3 + ( f ) 3 - ( | ) 3 is
(d) (0.2)3 - (0.3)3 + (0.1 )3
we know that,
a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - be - ca)
here, a = 0.2, b = 0.3, c = 0.1
a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - be - ca)
a3 + b3 + c3 = (a + b + c)(a2 + b2 + c2 - ab - be - ca) + 3abc
a3 + b3 + c3 = (0.2 - 0.3 + 0.1 )(a2 + b2 + c2 - ab - be - ca) + 3abc
a3 + b3 + c3 = (0)(a2 + b2 + c2 - ab - be - ca) + 3abc
a3 + b3 + c3 = 3abc
(0.2)3 - (0.3)3 + (0.1 )3 = 3abc
= 3(0.2)(-0.3)(0.1)
= -0.018
Hence, the value (0.2)3 - (0.3)3 + (0.1 )3 is 0.018