RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities

RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities

RD Sharma Solutions Class 9 Chapter 4 Algebraic Identities Ex 4.1

Question 1.
Evaluate each of the following using identities:
(i) (2x –\(\frac { 1 }{ x }\)) ²
(ii)  (2x + y) (2x – y)
(iii) (a ² b – b ² a) ²
(iv) (a – 0.1) (a + 0.1)
(v) (1.5.x ² – 0.3y ² ) (1.5x ² + 0.3y ² )
Solution:

Question 2.
Evaluate each of the following using identities:
(i) (399) ²
(ii) (0.98) ²
(iii) 991 x 1009
(iv) 117 x 83
Solution:

Question 3.
Simplify each of the following:

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.
If 9x ² + 25y ² = 181 and xy = -6, find the value of 3x + 5y.
Solution:
9x ² + 25y ² = 181, and xy = -6
(3x + 5y) ² = (3x) ² + (5y) ² + 2 x 3x + 5y
⇒ 9X ² + 25y ² + 30xy
= 181 + 30 x (-6)
= 181 – 180 = 1
= (±1 ) ²
∴ 3x + 5y = ±1

Question 8.
If 2x + 3y = 8 and xy = 2, find the value of 4X ² + 9y ² .
Solution:
2x + 3y = 8 and xy = 2
Now, (2x + 3y) ² = (2x) ² + (3y) ² + 2 x 2x x 3y
⇒  (8) ² = 4x ² + 9y ² + 12xy
⇒ 64 = 4X ² + 9y ² + 12 x 2
⇒ 64 = 4x ² + 9y ² + 24
⇒ 4x ² + 9y ² = 64 – 24 = 40
∴ 4x ² + 9y ² = 40

Question 9.
If 3x -7y = 10 and xy = -1, find the value of 9x ² + 49y ²
Solution:
3x – 7y = 10, xy = -1
3x -7y= 10
Squaring both sides,
(3x – 7y) ² = (10) ²
⇒ (3x) ² + (7y) ² – 2 x 3x x 7y = 100
⇒  9X ² + 49y ² – 42xy = 100
⇒  9x ² + 49y ² – 42(-l) = 100
⇒ 9x ² + 49y ² + 42 = 100
∴ 9x ² + 49y ² = 100 – 42 = 58

Question 10.
Simplify each of the following products:

Solution:

Question 11.

Solution:

Question 12.

Solution:

Question 13.
Simplify each of the following products:

Solution:

Question 14.
Prove that a ² + b ² + c ² – ab – bc – ca is always non-negative for all values of a, b and c.
Solution:

∵  The given expression is sum of these squares
∴ Its value is always positive Hence the given expression is always non-­negative for all values of a, b and c

Class 9 RD Sharma Solutions Chapter 4 Algebraic Identities Ex 4.2

Question 1.
Write the following in the expanded form:

Solution:

Question 2.
If a + b + c = 0 and a ² + b ² + c ² = 16, find the value of ab + be + ca.
Solution:
a + b+ c = 0
Squaring both sides,
(a + b + c) ² = 0
⇒ a ² + b ² + c ² + 2(ab + bc + ca) = 0
16 + 2(ab + bc + c) = 0
⇒ 2(ab + bc + ca) = -16
⇒  ab + bc + ca =-\(\frac { 16 }{ 2 }\) = -8
∴ ab + bc + ca = -8

Question 3.
If a ² + b ² + c ² = 16 and ab + bc + ca = 10, find the value of a + b + c.
Solution:
(a + b + c) ² = a ² + b ² + c ² + 2(ab + bc + ca)
= 16 + 2 x 10
= 16 + 20 = 36
= (±6) ²
∴ a + b + c = ±6

Question 4.
If a + b + c = 9 and ab + bc + ca = 23, find the value of a ² + b ² + c ² .
Solution:
(a + b + c) ² = a ² + b ² + c ² + 2(ab + bc + ca)
⇒ (9) ² = a ² + b ² + c ² + 2 x 23
⇒ 81= a ² + b ² + c ² + 46
⇒ a ² + b ² + c ² = 81 – 46 = 35
∴ a ² + b ² + c ² = 35

Question 5.
Find the value of 4x ² + y ² + 25z ² + 4xy – 10yz – 20zx when x = 4, y = 3 and z = 2.
Solution:
x = 4, y – 3, z = 2
⇒ 4x ² + y ² + 25z ² + 4xy – 10yz – 20zx
= (2x) ² + (y) ² + (5z) ² + 2 x2 x x y-2 x y x 5z – 2 x 5z x 2x
= (2x + y- 5z) ²
= (2 x 4 + 3- 5 x 2) ²
= (8 + 3- 10) ²
= (11 – 10) ²
= (1) ² = 1

Question 6.
Simplify:
(i)  (a + b + c) ² + (a – b + c) ²
(ii) (a + b + c) ² –  (a – b + c) ²
(iii) (a + b + c) ² +   (a – b + c) ² + (a + b – c) ²
(iv) (2x + p – c) ² – (2x – p + c) ²
(v) (x ² + y ² – z ² ) ² – (x ² – y ² + z ² ) ²
Solution:

Question 7.
Simplify each of the following expressions:

Solution:

Class 9 RD Sharma Solutions Chapter 4 Algebraic Identities Ex 4.3

Question 1.
Find the cube of each of the following binomial expressions:

Solution:

Question 2.
If a + b = 10 and ab = 21, find the value of a ³ + b ³ .
Solution:
a + b = 10, ab = 21
Cubing both sides,
(a + b) ³ = (10) ³
⇒ a ³ + 6 ³ + 3ab (a + b) = 1000
⇒  a ³ + b ³ + 3 x 21 x 10 = 1000
⇒  a ³ + b ³ + 630 = 1000
⇒  a ³ + b ³ = 1000 – 630 = 370
∴ a ³ + b ³ = 370

Question 3.
If a – b = 4 and ab = 21, find the value of a ³ -b ³ .
Solution:
a – b = 4, ab= 21
Cubing both sides,
⇒ (a – A) ³ = (4) ³
⇒ a ³ – b ³ – 3ab (a – b) = 64
⇒ a ³ -i ³ -3×21 x4 = 64
⇒  a ³ – 6 ³ – 252 = 64
⇒  a ³ – 6 ³ = 64 + 252 =316
∴ a ³ – b ³ = 316

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.
If 2x + 3y = 13 and xy = 6, find the value of 8x ³ + 21y ³ .
Solution:
2x + 3y = 13, xy = 6
Cubing both sides,
(2x + 3y) ³ = (13) ³
⇒ (2x) ³ + (3y) ³ + 3 x 2x x 3X2x + 3y) = 2197
⇒ 8x ³ + 27y ³ + 18xy(2x + 3y) = 2197
⇒ 8x ³ + 27y ³ + 18 x 6 x 13 = 2197
⇒ 8X ³ + 27y ³ + 1404 = 2197
⇒  8x ³ + 27y ³ = 2197 – 1404 = 793
∴ 8x ³ + 27y ³ = 793

Question 10.
If 3x – 2y= 11 and xy = 12, find the value of 27x ³ – 8y ³ .
Solution:
3x – 2y = 11 and xy = 12 Cubing both sides,
(3x – 2y) ³ = (11) ³
⇒  (3x) ³ – (2y) ³ – 3 x 3x x 2y(3x – 2y) =1331
⇒  27x ³ – 8y ³ – 18xy(3x -2y) =1331
⇒   27x ³ – 8y ³ – 18 x 12 x 11 = 1331
⇒  27x ³ – 8y ³ – 2376 = 1331
⇒  27X ³ – 8y ³ = 1331 + 2376 = 3707
∴ 2x ³ – 8y ³ = 3707

Question 11.
Evaluate each of the following:
(i)  (103) ³
(ii) (98) ³
(iii) (9.9) ³
(iv) (10.4) ³
(v) (598) ³
(vi) (99) ³
Solution:
We know that (a + bf = a ³ + b ³ + 3ab(a + b) and (a – b) ³ = a ³ – b ³ – 3 ab(a – b)
Therefore,
(i)  (103) ³ = (100 + 3) ³
= (100) ³ + (3) ³ + 3 x 100 x 3(100 + 3)    {∵ (a + b) ³ = a ³ + b ³ + 3ab(a + b)}
= 1000000 + 27 + 900 x 103
= 1000000 + 27 + 92700
= 1092727
(ii) (98) ³ = (100 – 2) ³
= (100) ³ – (2) ³ – 3 x 100 x 2(100 – 2)
= 1000000 – 8 – 600 x 98
= 1000000 – 8 – 58800
= 1000000-58808
= 941192
(iii) (9.9) ³ = (10 – 0.1) ³
= (10) ³ – (0.1) ³ – 3 X 10 X 0.1(10 – 0.1)
= 1000 – 0.001 – 3 x 9.9
= 1000 – 0.001 – 29.7
= 1000 – 29.701
= 970.299
(iv) (10.4) ³ = (10 + 0.4) ³
= (10) ³ + (0.4) ³ + 3 x 10 x 0.4(10 + 0.4)
= 1000 + 0.064 + 12(10.4)
= 1000 + 0.064 + 124.8 = 1124.864
(v) (598) ³ = (600 – 2) ³
= (600) ³ – (2) ³ – 3 x 600 x 2 x (600 – 2)
= 216000000 – 8 – 3600 x 598
= 216000000 – 8 – 2152800
= 216000000 – 2152808
= 213847192
(vi) (99) ³ = (100 – 1) ³
= (100) ³ – (1) ³ – 3 x 100 x 1 x (100 – 1)
= 1000000 – 1 – 300 x 99
= 1000000 – 1 – 29700
= 1000000 – 29701
= 970299

Question 12.
Evaluate each of the following:
(i)  111 ³ – 89 ³
(ii) 46 ³ + 34 ³
(iii) 104 ³ + 96 ³
(iv) 93 ³ – 107 ³
Solution:
We know that a ³ + b ³ = (a + bf – 3ab(a + b) and a ³ – b ³ = (a – bf + 3 ab(a – b)
(i) 111 ³ – 89 ³
= (111 – 89) ³ + 3 x ill x 89(111 – 89)
= (22) ³ + 3 x 111 x 89 x 22
= 10648 + 652014 = 662662
(OR)
(a + b) ³ – (a – b) ³ = 2(b ³ + 3a ² b)
= 111 ³ – 89 ³ = (100 + 11) ³ – (100 – 11) ³
= 2(11 ³ + 3 x 100 ² x 11]
= 2(1331 + 330000]
= 331331 x 2 = 662662
(a + b) ³ + (a- b) ³ = 2(b ³ + 3ab ² )
(ii) 46 ³ + 34 ³ = (40 + 6) ³ + (40 – 6) ³
= 2[(40) ³ + 3 x 40 x 6 ² ]
= 2[64000 + 3 x 40 x 36]
= 2[64000 + 4320]
= 2 x 68320 = 136640
(iii) 104 ³ + 96 ³ = (100 + 4) ³ + (100 – 96) ³
= 2 [a ³ + 3 ab ² ]
= 2[(100) ³ + 3 x 100 x (4) ² ]
= 2[ 1000000 + 300 x 16]
= 2[ 1000000 + 4800]
= 1004800 x 2 = 2009600
(iv) 93 ³ – 107 ³ = -[(107) ³ – (93) ³ ]
= -[(100 + If – (100 – 7) ³ ]
= -2[b ³ + 3a ² b)]
= -2[(7) ³ + 3(100) ² x 7]
= -2(343 + 3 x 10000 x 7]
= -2[343 + 210000]
= -2[210343] = -420686

Question 13.

Solution:

Question 14.
Find the value of 27X ³ + 8y ³ if
(i) 3x + 2y = 14 and xy = 8
(ii) 3x + 2y = 20 and xy = \(\frac { 14 }{ 9 }\)
Solution:

Question 15.
Find the value of 64x ³ – 125z ³ , if 4x – 5z = 16 and xz = 12.
Solution:
4x – 5z = 16, xz = 12
Cubing both sides,
(4x – 5z) ³ = (16) ³
⇒ (4x) ³ – (5y) ³ – 3 x 4x x 5z(4x – 5z) = 4096
⇒ 64x ³ – 125z ³ – 3 x 4 x 5 x xz(4x – 5z) = 4096
⇒  64x ³ – 125z ³ – 60 x 12 x 16 = 4096
⇒ 64x ³ – 125z ³ – 11520 = 4096
⇒  64x ³ – 125z ³ = 4096 + 11520 = 15616

Question 16.

Solution:

Question 17.
Simplify each of the following:

Solution:

Question 18.


Solution:

Question 19.

Solution:

RD Sharma Mathematics Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.4

Question 1.
Find the following products:
(i) (3x + 2y) (9X ² – 6xy + Ay ² )
(ii) (4x – 5y) (16x ² + 20xy + 25y ² )
(iii) (7p ⁴ + q) (49p ⁸ – 7p ⁴ q + q ² )

Solution:

Question 2.
If x = 3 and y = -1, find the values of each of the following using in identity:

Solution:

Question 3.
If a + b = 10 and ab = 16, find the value of a ² – ab + b ² and a ² + ab + b ² .
Solution:
a + b = 10, ab = 16 Squaring,
(a + b) ² = (10) ²
⇒ a ² + b ² + lab = 100
⇒ a ² + b ² + 2 x 16 = 100
⇒  a ² + b ² + 32 = 100
∴ a ² + b ² = 100 – 32 = 68
Now, a ² – ab + b ² = a ² + b ² – ab = 68 – 16 = 52
and a ² + ab + b ² = a ² + b ² + ab = 68 + 16 = 84

Question 4.
If a + b = 8 and ab = 6, find the value of a ³ + b ³ .
Solution:
a + b = 8, ab = 6
Cubing both sides,
(a + b) ³ = (8)3
⇒ a ³ + b ³ + 3 ab{a + b) = 512
⇒  a ³ + b ³ + 3 x 6 x 8 = 512
⇒  a ³ + b ³ + 144 = 512
⇒  a ³ + b ³ = 512 – 144 = 368
∴ a ³ + b ³ = 368

Question 5.
If a – b = 6 and ab = 20, find the value of a ³ -b ³ .
Solution:
a – b = 6, ab = 20
Cubing both sides,
(a – b)3 = (6) ³
⇒  a ³ – b ³ – 3ab(a – b) = 216
⇒  a ³ – b ³ – 3 x 20 x 6 = 216
⇒  a ³ – b ³ – 360 = 216
⇒  a ³ -b ³ = 216 + 360 = 576
∴ a ³ – b ³ = 576

Question 6.
If x = -2 and y = 1, by using an identity find the value of the following:

Solution:

RD Sharma Solutions Class 9 Chapter 4 Algebraic Identities Ex 4.5

Question 1.
Find the following products:
(i) (3x + 2y + 2z) (9x ² + 4y ² + 4z ² – 6xy – 4yz – 6zx)
(ii) (4x -3y + 2z) (16x ² + 9y ² + 4z ² + 12xy + 6yz – 8zx)
(iii) (2a – 3b – 2c) (4a ² + 9b ² + 4c ² + 6ab – 6bc + 4ca)
(iv) (3x -4y + 5z) (9x ² + 16y ² + 25z ² + 12xy- 15zx + 20yz)
Solution:
(i) (3x + 2y + 2z) (9x ² + 4y ² + 4z ² – 6xy – 4yz – 6zx)
= (3x + 2y + 2z) [(3x) ² + (2y) ² + (2z) ² – 3x x 2y + 2y x 2z + 2z x 3x]
= (3x) ³ + (2y) ³ + (2z) ³ – 3 x 3x x 2y x 2z
= 27x ³ + 8y ³ + 8Z ³ – 36xyz
(ii) (4x – 3y + 2z) (16x ² + 9y ² + 4z ² + 12xy + 6yz – 8zx)
= (4x -3y + 2z) [(4x) ² + (-3y) ² + (2z) ² – 4x x (-3y + (3y) x (2z) – (2z x 4x)]
= (4x) ³ + (-3y) ³ + (2z) ³ – 3 x 4x x (-3y) x (2z)
= 64x ³ – 27y ³ + 8z ³ + 72xyz
(iii) (2a -3b- 2c) (4a ² + 9b ² + 4c ² + 6ab – 6bc + 4ca)
= (2a -3b- 2c) [(2a) ² + (3b) ² + (2c) ² – 2a x (-3b) – (-3b) x (-2c) – (-2c) x 2a]
= (2a) ³ + (3b) ³ + (-2c) ³ -3x 2a x (-3 b) (-2c)
= 8a ³ – 21b ³ -8c ³ – 36abc
(iv) (3x – 4y + 5z) (9x ² + 16y ² + 25z ² + 12xy – 15zx + 20yz)
= [3x + (-4y) + 5z] [(3x) ² + (-4y) ² + (5z) ² – 3x x (-4y) -(-4y) (5z) – 5z x 3x]
= (3x) ³ + (-4y) ³ + (5z) ³ – 3 x 3x x (-4y) (5z)
= 27x ³ – 64y ³ + 125z ³ + 180xyz

Question 2.
Evaluate:

Solution:

Question 3.
If x + y + z = 8 and xy + yz+ zx = 20, find the value of x ³ + y ³ + z ³ – 3xyz.
Solution:
We know that
x ³ + y ³ + z ³ – 3xyz = (x + y + z) (x ² + y ² + z ² -xy -yz – zx)
Now, x + y + z = 8
Squaring, we get
(x + y + z) ² = (8) ²
x ² + y ² + z ² + 2(xy + yz + zx) = 64
⇒ x ² + y ² + z ² + 2 x 20 = 64
⇒  x ² + y ² + z ² + 40 = 64
⇒  x ² + y ² + z ² = 64 – 40 = 24
Now,
x ³ + y ³ + z ³ – 3xyz = (x + y + z) [x ² + y ² + z ² – (xy + yz + zx)]
= 8(24 – 20) = 8 x 4 = 32

Question 4.
If a +b + c = 9 and ab + bc + ca = 26, find the value of a ³ + b ³ + c ³ – 3abc.
Solution:
a + b + c = 9, ab + be + ca = 26
Squaring, we get
(a + b + c) ² = (9) ²
a ² + b ² + c ² + 2 (ab + be + ca) = 81
⇒  a ² + b ² + c ² + 2 x 26 = 81
⇒ a ² + b ² + c ² + 52 = 81
∴  a ² + b ² + c ² = 81 – 52 = 29
Now, a ³ + b ³ + c ³ – 3abc = (a + b + c) [(a ² + b ² + c ² – (ab + bc + ca)]
= 9[29 – 26]
= 9 x 3 = 27

Question 5.
If a + b + c = 9, and a ² + b ² + c ² = 35, find the value of a ³ + b ³ + c ³ – 3abc.
Solution:
a + b + c = 9
Squaring, we get
(a + b + c) ² = (9) ²
⇒  a ² + b ² + c ² + 2 (ab + be + ca) = 81
⇒  35 + 2(ab + bc + ca) = 81
2(ab + bc + ca) = 81 – 35 = 46
∴  ab + bc + ca = \(\frac { 46 }{ 2 }\) = 23
Now, a ³ + b ³ + c ³ – 3abc
= (a + b + c) [a ² + b ² + c ² – (ab + bc + ca)]
= 9[35 – 23] = 9 x 12 = 108

Class 9 Maths Chapter 4 Algebraic Identities RD Sharma Solutions VSAQS

Question 1.

Solution:

Question 2.

Solution:

Question 3.
If a + b = 7 and ab = 12, find the value of a ² + b ² .
Solution:
a + b = 7, ab = 12
Squaring both sides,
(a + b) ² = (7) ²
⇒  a ² + b ² + 2ab = 49
⇒  a ² + b ² + 2 x 12 = 49
⇒ a ² + b ² + 24 = 49
⇒ a ² + b ² = 49 – 24 = 25
∴ a ² + b ² = 25

Question 4.
If a – b = 5 and ab = 12, find the value of a ² + b ² .
Solution:
a – b = 5, ab = 12
Squaring both sides,
⇒ (a – b) ² = (5) ²
⇒  a ² + b ² – 2ab = 25
⇒  a ² + b ² – 2 x 12 = 25
⇒  a ² + b ² – 24 = 25
⇒  a ² + b ² = 25 + 24 = 49
∴ a ² + b ² = 49

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Algebraic Identities Class 9 RD Sharma Solutions MCQS

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.
If a + b + c = 9 and  ab + bc + ca = 23, then a ² + b ² + c ² =
(a) 35
(b) 58
(c) 127
(d) none of these
Solution:
a + b + c = 9, ab + bc + ca = 23
Squaring,
(a + b+ c) = (9) ²
a ² + b ² + c ² + 2 (ab + bc + ca) = 81
⇒ a ² + b ² + c ² + 2 x 23 = 81
⇒  a ² + b ² + c ² + 46 = 81
⇒  a ² + b ² + c ² = 81 – 46 = 35   (a)

Question 9.
(a – b) ³ + (b – c) ³ + (c – a) ³ =
(a) (a + b + c) (a ² + b ² + c ² – ab – bc – ca)
(d) (a -b)(b- c) (c – a)
(c) 3(a – b) (b – c) (c – a)
(d) none of these
Solution:
(a – b) ³ + (b- c) ³ + (c- a) ³
∵ a – b + b – c + c – a = 0
∴ (a – b) ³ + (b – c) ³ + (c – a) ³
= 3 (a -b)(b- c) (c – a)        (c)

Question 10.

Solution:

Question 11.
If a – b = -8 and ab = -12 then a ³ – b ³ =
(a) -244
(b) -240
(c) -224
(d) -260
Solution:
a – b = -8, ab = -12
(a – b) ³ = a ³ – b ³ – 3ab (a – b)
(-8) ³ = a ³ – b ³ – 3 x (-12) (-8)
-512 = a ³ -b ³ – 288
a ³ – b ³ = -512 + 288 = -224      (c)

Question 12.
If the volume of a cuboid is 3x ² – 27, then its possible dimensions are
(a) 3, x ² , -27x
(b) 3, x – 3, x + 3
(c) 3, x ² , 27x
(d) 3, 3, 3
Solution:
Volume = 3x ² -27 = 3(x ² – 9)
= 3(x + 3) (x – 3)
∴ Dimensions are   = 3, x – 3, x   +  3        (b)

Question 13.
75 x 75 + 2 x 75 x 25    + 25 x 25 is equal to
(a) 10000
(b) 6250
(c) 7500
(d) 3750
Solution:

Question 14.
(x – y) (x + y)(x ² + y ² ) (x ⁴ + y ⁴ ) is equal to
(a) x ¹⁶ – y ¹⁶
(b) x ⁸ – y ⁸
(c) x ⁸ + y ⁸
(d) x ¹⁶ + y ¹⁶
Solution:

Question 15.

Solution:

Question 16.

Solution:

Question 17.

Solution:

Question 18.

Solution:

Question 19.
If a ² + b ² + c ² – ab – bc – ca = 0, then
(a) a + b = c
(b) b + c = a
(c) c + a = b
(d) a = b = c
Solution:
a ² + b ² + c ² – ab – bc – ca = 0
2 {a ² + b ² + c ² – ab – be – ca) = 0 (Multiplying by 2)
⇒  2a ² + 2b ² + 2c ² – 2ab – 2bc – 2ca = 0
⇒  a ² + b ² – 2ab + b ² + c ² – 2bc + c ² + a ² – 2ca = 0
⇒  (a – b) ² + (b – c) ² + (c – a) ² = 0
(a – b) ² = 0, then a – b = 0
⇒ a = b
Similarly, (b – c) ² = 0, then
b-c = 0
⇒ b = c
and (c – a) ² = 0, then c-a = 0
⇒ c = a
∴ a = b – c           (d)

Question 20.

Solution:

Question 21.


Solution:

Question 22.
If a + b + c = 9 and ab + bc + ca = 23, then a ³ + b ³ + c ³ – 3 abc =
(a) 108
(b) 207
(c) 669
(d) 729
Solution:
a ³ + b ³ + c ³ – 3abc
= (a + b + c) [a ² + b ² + c ² – (ab + bc + ca)
Now, a + b + c = 9
Squaring,
a ² + b ² + c ² + 2 (ab + be + ca) = 81
⇒  a ² + b ² + c ² + 2 x 23 = 81
⇒  a ² + b ² + c ² + 46 = 81
⇒  a ² + b ² + c ² = 81 – 46 = 35
Now, a ³ + b ³ + c ³ – 3 abc = (a + b + c) [(a ² + b ² + c ² ) – (ab + bc + ca)
= 9[35 -23] = 9 x 12= 108                     (a)

Question 23.

Solution:

Question 24.
The product (a + b) (a – b) (a ² – ab + b ² ) (a ² + ab + b ² ) is equal to
(a) a ⁶ +   b ⁶
(b) a ⁶ – b ⁶
(c) a ³ – b ³
(d) a ³ + b ³
Solution:
(a + b) (a – b) (a ² – ab + b ² ) (a ² + ab +b ² )
= (a + b) (a ² -ab + b ² ) (a-b) (a ² + ab + b ² )
= (a ³ + b ³ ) (a ³ – b ³ )
= (a ³ ) ² – (b ³ ) ² = a ⁶ – b ⁶ (b)

Question 25.
The product (x ² – 1) (x ⁴ + x ² + 1) is equal to
(a) x ⁸ –   1
(b) x ⁸ + 1
(c) x ⁶ –   1
(d) x ⁶ + 1
Solution:
(x ² – 1) (x ⁴ + x ² + 1)
= (x ² ) ³ – (1) ³ = x ⁶ – 1                            (c)

Question 26.

Solution:

Question 27.

Solution:

RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities

• RD Sharma Class 9th Solutions Chapter 4 Algebraic Identities Exercise 4.1
  RD Sharma Class 9th Solutions Chapter 4 Algebraic Identities Exercise 4.2
  RD Sharma Class 9th Solutions Chapter 4 Algebraic Identities Exercise 4.3
• RD Sharma Class 9th Solutions Chapter 4 Algebraic Identities Exercise 4.4
• RD Sharma Class 9th Solutions Chapter 4 Algebraic Identities Exercise 4.5

RD Sharma Class 9 Solutions