RS Aggarwal Class 8 Maths Solutions

RS Aggarwal Class 8 Solutions Chapter 7 Factorisation Ex 7C

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7C.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7A
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7B
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7C
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7D
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7E

Question 1.
Solution:
x² + 8x + 16
= (x)² + 2 × x × 4 + (4)²
= (x + 4)²

Question 2.
Solution:
x² + 14x + 49
= (x)² + 2 × x × 7 + (7)²
= (x + 7)² Ans.

Question 3.
Solution:
1 + 2x + x²
= (1)² + 2 × 1 × x + (x)²
= (1 + x)² Ans.

Question 4.
Solution:
9 + 6z + z²
= (3)² + 2 x 3 x z + (z)²
= (3 + z)² Ans.

Question 5.
Solution:
x² + 6ax + 9a²
= (x)² + 2 × x × 3a + (3a)²
= (x + 3a)² Ans.

Question 6.
Solution:
4y² + 20y + 25
= (2y)² + 2 x 2y x 5 + (5)²
= (2y² + 5)²
{ ∵ a² + 2ab + b² = (a + b)²}

Question 7.
Solution:
36a² + 36a + 9
= 9 [4a² + 4a + 1]
= 9 [(2a)² + 2 x 2a x 1 + (1)²]
= 9 [2a + 1]²

Question 8.
Solution:
9m² + 24m + 16
= (3m)² + 2 x 3m x 4 + (4)²
= (3m + 4)²
{ ∵ a² + 2ab + b² = (a + b)²}

Question 9.
Solution:
z² + z + \(\\ \frac { 1 }{ 4 } \)
= (z)² + 2 x z x \(\\ \frac { 1 }{ 2 } \) + \({ \left( \frac { 1 }{ 2 } \right) }^{ 2 }\)
= \({ \left( z+\frac { 1 }{ 2 } \right) }^{ 2 }\)

Question 10.
Solution:
49a² + 84ab + 36b²
= (7a)² + 2 x 7a x 6b + (6b)²
{ ∵ a² + 2ab + b² = (a + b)²}
= (7a + 6b)²

Question 11.
Solution:
p² – 10p + 25
= (p)² – 2 x p x 5 + (5)²
= (p – 5)²
{ ∵ a² – 2ab + b² = (a – b)²}

Question 12.
Solution:
121a² – 88ab + 16b²
= (11a)² – 2 x 11a x 4b + 4(b)²
= (11a – 4b)²

Question 13.
Solution:
1 – 6x + 9x²
= (1)² – 2 x 1 x 3x + (3x)²
= (1 – 3x)²
{ ∵ a² – 2ab + b² = (a – b)²}

Question 14.
Solution:
9y² – 12y + 4
= (3y)² – 2 x 3y x 2 + (2)²
{ ∵ a² – 2ab + b² = (a – b)²}
= (3y – 2)²

Question 15.
Solution:
16x² – 24x + 9
= (4x)² – 2 x 4x x 3 + (3)²
= (4x – 3)² Ans.

Question 16.
Solution:
m² – 4mn + 4n²
= (m)² -2 x m x 2n + (2n)²
= (m – 2n)² Ans.

Question 17.
Solution:
a²b² – 6abc + 9c²
= (ab)² – 2 x ab x 3c + (3c)²
= (ab – 3c)² Ans.

Question 18.
Solution:
m⁴ + 2m²n² + n⁴
= (m²)² + 2m²n² + (n²)²
= (m² + n²)²
{ ∵ a² + 2ab + b² = (a + b)²}

Question 19.
Solution:
(l + m)² – 4lm
= l² + m² + 2lm – 4lm
= l² + m² – 2lm
= l² – 2lm + m²
= (l – m)²
{ ∵ a² – 2ab + b² = (a – b)}

Hope given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 7 Factorisation Ex 7B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7B.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7A
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7B
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7C
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7D
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7E

Question 1.
Solution:
x² – 36
= (x)² – (6)² { ∵ a² – b² = (a + b) (a – b)}
= (x + 6) (x – 6) Ans.

Question 2.
Solution:
4a² – 9
= (2a)² – (3)²
= (2a + 3) (2a – 3)
{ ∵ a² – b² = (a + b) (a – b)}

You can also Download NCERT Solutions for Class 8 English to help you to revise complete Syllabus and score more marks in your examinations.

Question 3.
Solution:
81 – 49x²
= (9)² – (7x)²
= (9 + 7x) (9 – 7x) { ∵ a² – b² = (a + b) (a – b)}

Question 4.
Solution:
= (2x)² – (3y)²
= (2x + 3y)(2x-3y)
{∵ a² – b² = (a + b) (a – b)}

Question 5.
Solution:
Using a² – b²
= (a + b) (a – b)
= 16a² – 225b²
= (4a)² – (15b)²
= (4a + 15b) (4b – 5b)

Question 6.
Solution:
Using a² – b²
= (a + b) (a – b)
= (3ab)² – (5)²
= (3ab + 5) (3ab – 5)

Question 7.
Solution:
Using a² – b² = (a + b) (a – b)
16a² – 144 = (4a)² = (12)²
= (4a + 12) (4a – 12)
= 4 (a + 3) x 4 (a – 3)
= 16 (a + 3) (a – 3)

Question 8.
Solution:
63a² – 112b²
= 7 (9a² – 16b²)
= 7 [(3a)² – (4b)²
= 7 (3a + 4b) (3a – 4b)

Question 9.
Solution:
20a² – 45b²
= 5 {4a² – 9b²}
= 5{(2a)² – (3b)²)
= 5(2a + 3b) (2a – 3b) Ans.

Question 10.
Solution:
12x² – 27
= 3(4x² – 9)
= 3{(2x)² – (3)²}
= 3(2x + 3) (2x – 3) Ans.

Question 11.
Solution:
x³ – 64x
= x(x² – 64)
= x{(x)² – (8)²}
= x(x + 8) (x – 8) Ans.

Question 12.
Solution:
16x⁵ – 144x³
= 16x³ [x² – 9]
= 16x³ [(x)² – (3)²]
= 16x³ (x + 3) (x – 3)

Question 13.
Solution:
3x⁵ – 48x³
= 3x³ {x² – 16}
= 3x³{(x)² – (4)²}
= 3x³ (x + 4) (x – 4) Ans.

Question 14.
Solution:
16p³ – 4p
= 4p [4p² – 1]
= 4p ((2p)² – (1)²]
= 4p(2p + 1)(2p – 1)

Question 15.
Solution:
63a²b² – 7
= 7(9a²b² – 1)
= 7{(3ab)² – (1)²)
= 7(3ab + 1) (3ab – 1) Ans.

Question 16.
Solution:
1 – (b – c)²
= (1)² – (b – c)²
= (1 + b + c) (1 – b + c) Ans.
{ ∵ a² – b² = (a + b) (a – b)}

Question 17.
Solution:
(2a + 3b)² – 16c²
= (2a + 3b)² – (4c)²
=(2a + 3b + 4c)(2a + 3b – 4c)Ans.
{ ∵ a² – b² = (a + b)(a – b)}

Question 18.
Solution:
(l + m)² – (l – m)²
= (l + m + l – m)(l + m – l + m)
{ ∵ a² – b² = (a + b)(a – b)}
= 2l x 2m = 4lm

Question 19.
Solution:
(2x + 5y)² – (1)²
=(2x + 5y + 1)(2x + 5y – 1)
{ ∵ a² – b² = (a + b)(a – b)}

Question 20.
Solution:
36c² – (5a + b)²
= (6c)² – (5a + b)²
{ ∵ a² – b² = (a + b)(a – b)}
= (6c + 5a + b)(6c – 5a – b)

Question 21.
Solution:
(3x – 4y)² – 25z²
= (3x – 4y)² – (5z)²
= (3x – 4y + 5z) (3x – 4y – 5z) Ans.

Question 22.
Solution:
x² – y² – 2y – 1
= x² – (y² + 2y + 1)
= (x)² – (y + 1)²
= (x + y + 1)(x – y – 1)Ans.

Question 23.
Solution:
25 – a² – b² – 2ab
= 25 – (a² + b² + 2ab)
= (5)² – (a + b)²
= (5 + a + b)(5 – a – b)Ans.

Question 24.
Solution:
25a² – 4b² + 28bc – 49c²
= 25a² – [4b² – 28bc + 49c²]
{ ∵ a² – 2ab + b² = (a – b)²}
= (5a)² – [(2b)² – 2 x 2b x 7c + (7c)²]
= (5a)² – (2b – 7c)²
{ ∵ (a² – b² = (a + b)(a – b)}
= (5a + 2b – 7c) (5a – 2b + 7c)

Question 25.
Solution:
9a² – b² + 4b – 4
= 9a² – (b² – 4b + 4)
= (3a)² – [(b)² – 2 x b x 2 + (2)²]
= (3a)² – (b – 2)²
{ ∵ a² – 2ab + b² = (a – b)²}
= (3a + b – 2)(3a – b + 2)
{ ∵ a² – b² = (a + b)(a – b)}

Question 26.
Solution:
(10)² – (x – 5)²
= (10)² – (x – 5)²
= (10 + x – 5)(10 – x + 5)
= (5 + x) (15 – x) Ans.

Question 27.
Solution:
{(405)² – (395)²}
= (405)² – (395)²
= (405 + 395) (405 – 395)
{ ∵ a² – b²(a + b) (a – b)}
= 800 x 10 = 8000

Question 28.
Solution:
(7.8)² – (2.2)²
= (7.8 + 2.2) (7.8 – 2.2)
= 10.0 x 5.6
= 56 Ans.

Hope given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 7 Factorisation Ex 7A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7A.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7A
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7B
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7C
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7D
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7E

Factorize :

Question 1.
Solution:
(i) 12x + 15 = 3(4x + 5) {HCF of 12 and 15 = 3)
(ii) 14m – 21 = 7(2m – 3) {HCF of 14 and 21 = 7)
(iii) 9n – 12n² = 3n(3 – 4n) Ans.{HCF of 9 and 12 = 3}

Question 2.
Solution:
(i) 16a² – 24ab = 8a(2a – 3b) { HCF of 16 and 24 = 8)
(ii) 15ab² – 20a²b = 5ab(3b – 4a) {HCF of 15 and 20 5}
(iii) 12x²y³ – 21x³y²= 3x²y²(4y – 7x) Ans. {HCF of 12 and 21 = 3)

Question 3.
Solution:
(i) 24x³ – 36x²y = 12x²(2x – 3y) {HCF of 24 and 36 = 12}
(ii) 10x³ – 15x² = 5x²(2x – 3) {HCF of 10, 15 = 5}
(iii) 36x³y – 60x²y³z = 12x²y (3x – 5y²z) {HCF of 36 and 60 = 12}

Question 4.
Solution:
(i)9x³ – 6x² + 12x
= 3x(3x² – 2x + 4) {HCF of 9, 6, 12 = 3)
(ii)8x² – 72xy + 12x
= 4x(2x – 18y + 3) (HCF of 8, 72, 124)
(iii)18a³b³ – 27a²b³ + 36a³b²
= 9a²b²(2ab – 3b + 4a) Ans.{HCF of 10, 27, 36 = 9)

Question 5.
Solution:
(i) 14x³ + 21x⁴y – 28x²y²
= 7x² (2x + 3x²y – 4y²) {HCF of 14, 21 28 = 7)
(ii) – 5 – 10t + 20t²
= – 5(1 + 2t – 4t²) Ans. {HCF of 5, 10, 20=5)

Question 6.
Solution:
(i) x(x + 3) + 5(x + 3)
=(x + 3)(x + 5)
(ii) 5x(x – 4) – 7(x – 4)
= (x – 4) (5x – 7)
(iii) 2m (1 – n) + 3(1 – n)
= (1 – n) (2m + 3) Ans.

Question 7.
Solution:
6a(a – 2b) + 5b(a – 2b)
= (a – 2b) (6a + 5b) Ans.

Question 8.
Solution:
x³(2a – b) + x²(2a – b)
=x²(2a – b)(x + 1)Ans.

Question 9.
Solution:
9a(3a – 5b) – 12a²(3a – 5b)
= 3a (3a – 5b) (3 – 4a) Ans.

Question 10.
Solution:
(x + 5)² – 4(x + 5)
= (x + 5)(x + 5 – 4)
=(x + 5)(x + 1)Ans.

Question 11.
Solution:
3(a – 2b)² – 5 (a – 2b)
= (a – 2b) {3(a – 2b) – 5}
= (a – 2b) (3a – 6b – 5) Ans.

Question 12.
Solution:
2a + 6b – 3 (a + 3b)²
= 2(a + 3b) – 3(a + 3b)²
= (a + 3b){ 2 – 3 (a + 3b)}
= (a + 3b) (2 – 3a – 9b) Ans.

Question 13.
Solution:
16(2p – 3q)² – 4 (2p – 3q)
= 4(2p – 3q) {4(2p – 3q) – 1}
= 4(2p – 3q) (8p – 12q – 1) Ans.

Question 14.
Solution:
x(a – 3) + y (3 – a)
= x(a – 3) – y(a – 3)
= (a – 3) (x – y) Ans.

Question 15.
Solution:
12(2x – 3y)² – 16(3y – 2x)
= 12(2x – 3y)² + 16(2x – 3y)
= 4(2x – 3y) {3(2x – 3y) + 4}
= 4(2x – 3y) (6x – 9y + 4) Ans.

Question 16.
Solution:
(x + y)(2x + 5) – (x + y)(x + 3)
= (x + y)(2 + 5 – x – 3)
= (x + y)(x + 2) Ans.

Question 17.
Solution:
ar + br + at + bt
= r(a + b) + t(a + b)
= (a + b) (r + t) Ans.

Question 18.
Solution:
x² – ax – bx + ab
= x(x – a) – b(x – a)
= (x – a)(x – b)Ans.

Question 19.
Solution:
ab² – bc² – ab + c²
= ab² – ab – bc² + c²
= ab(b – 1) – c²(b – 1)
= (b – 1) (ab – c²) Ans.

Question 20.
Solution:
x² – xz + xy – yz
= x(x – z) + y(x – z)
= (x – z)(x + y)Ans.

Question 21.
Solution:
6ab – b² + 12ac – 2bc
6ab + 12ac – b² – 2bc
= 6a(b + 2c) – b(b + 2c)
= (b + 2c) (6a – b) Ans.

Question 22.
Solution:
(x – 2y)² + 4x – 8y
= (x – 2y)² + 4(x – 2y)
= (x – 2y)(x – 2y + 4)Ans.

Question 23.
Solution:
y² – xy(1 – x) – x³
y² – xy + x²y – x²
= y(y – x) + x² (y – x)
= (y – x)(y + x²)Ans.

Question 24.
Solution:
(ax + by)² + (bx – ay)²
= a²x² + b²y² + 2abxy + b²x² + a²y² – 2abxy
= a²x² + b²y² + b²x² + a²y²
= a²x² + b²x² + a²y² + b²y²
= x²(a² + b²) + y²(a² + b²)
= (a² + b²) (x² + y²) Ans.

Question 25.
Solution:
ab² + (a – 1)b – 1
= ab² + ab – b – 1
= ab(b + 1) – 1(b + 1)
= (b + 1) (ab – 1) Ans.

Question 26.
Solution:
x³ – 3x² + x – 3
= x²(x – 3) + 1(x – 3)
(x – 3)(x² + 1)Ans.

Question 27.
Solution:
ab (x² + y²) – xy (a² + b²)
= abx² + aby² + xya² – xyb²
= abx² – xya² – xyb² + aby2
=ax(bx – ay) – by(bx – ay)
= (bx – ay) (ax – by) Ans.

Question 28.
Solution:
x² – x (a + 2b) + 2ab
= x² – xa – 2bx + 2ab
= x(x – a) – 2b(x – a)
= (x – a)(x – 2b)

Hope given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 6 Operations on Algebraic Expressions Ex 6E

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6E.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6A
• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6B
• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6C
• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6D
• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6E

Objective questions :
Tick the correct answer in each of the following :

Question 1.
Solution:

Question 2.
Solution:
(3q + 7p² – 2r³ + 4) – (4p² – 2q + 7r³ – 3)
= 3q + 7p² – 2r³ + 4 – 4p² + 2q – 7r³ + 3
= 5q + 3p² – 9r³ + 7 = 3p² + 5q – 9r³ + 7 (d)

Question 3.
Solution:
(x + 5) (x – 3) = x² + (5 – 3) x + 5 X ( – 3)
= x² + 2x – 15 (d)

Question 4.
Solution:
(2x + 3) (3x – 1)
= 6x² – 2x + 9x – 3
= 6x² + 7x – 3 (b)

Question 5.
Solution:
(x + 4) (x + 4)
= x² + (4 + 4) x + 4 X 4
= x² + 8x + 16 (c)

Question 6.
Solution:
(x – 6) (x – 6)
= x² + ( – 6 – 6) x + ( – 6) ( – 6)
= x² – 12x + 36 (d)

Question 7.
Solution:
(2x + 5) (2x – 5)
= (2x)² – (5)²
= 4x² – 25

Question 8.
Solution:

Question 9.
Solution:
(2x² + 3x + 1) ÷ (x + 1)

= 2x +1 (b)

Question 10.
Solution:
(x² – 4x + 4) ÷ (x – 2)

= x – 2 (a)

Question 11.
Solution:
(a + 1) (a – 1) (a² + 1)
= (a² – 1) (a² + 1)
= a⁴ (c)

Question 12.
Solution:

Question 13.
Solution:
\(\left( { x }+\frac { 1 }{ { x } } \right) =5\)
Squaring on both sides

Question 14.
Solution:
\(\left( { x }-\frac { 1 }{ { x } } \right) =6\)
Squaring on both sides

Question 15.
Solution:
(82)² – (18)²
= (82 + 18) (82 – 18)
= 100 x 64
= 6400 (c)

Question 16.
Solution:
(197 x 203)
= (200 – 3) (200 + 3)
= (200)² – (3)²
= 40000 – 9
= 39991 (a)

Question 17.
Solution:
a + b = 12, ab = 14
a² + b²
= (a + b)² – 2ab
= (12)² – 2 x 14
= 144 – 28
= 16 (b)

Question 18.
Solution:
a – b = 7, ab = 9
a² + b²
= (a – b)² + 2ab
= (7)² + 2 x 9
= 49 + 18
= 67 (a)

Question 19.
Solution:
x = 10
4x² + 20x + 25
= (2x)² + 2 x 2x x 5 + (5)²
= (2x + 5)²
= (2 x 10 + 5)
= (20 + 5)²
= (25)²
= 625 (c)

Hope given RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6E are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 6 Operations on Algebraic Expressions Ex 6D

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6D.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6A
• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6B
• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6C
• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6D
• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6E

Question 1.
Solution:
(i)(x + 6)(x + 6)
= (x + 6)²

You can also Download NCERT Solutions for Class 8 English to help you to revise complete Syllabus and score more marks in your examinations.

Question 2.
Solution:
(i) (x – 4)(x – 4)
= (x – 4)²

Question 3.
Solution:
(i)(8a + 3b)²
= (8a)² + 2 x 8a x 3b + (3b)²

Question 4.
Solution:
(i)(x + 3)(x – 3)
= (x)² – (3)²

Question 5.
Solution:
(i) (54)²
= (50 + 4)²

Question 6.
Solution:
(i) (69)²
= (70 – 1)²

Question 7.
Solution:
(i)(82)² – (18)²
= (82 – 18)(82 – 18)

Question 8.
Solution:
9x² + 24x + 16
= (3x)² + 2 x 3x x 4 + (4)²
= (3x + 4)²
= (3 x 12 + 4)²
= (36 + 4)²
= (40)²
= 1600 Ans.

Question 9.
Solution:
64x² + 81y² + 144xy = (8x)² + (9y)² + 2 x 8x x 9y
= (8x + 9y)²
= \({ \left( 8\times 11+9\times \frac { 4 }{ 3 } \right) }^{ 2 }\)
= (88 + 12)²
= (100)²
= 10000 Ans.

Question 10.
Solution:
(36x² + 25y² – 60xy)
= (6x)² + (5y)² – 2 x 6x x 5y
= (6x – 5y)²
= \({ \left( 6\times \frac { 2 }{ 3 } -5\times \frac { 1 }{ 5 } \right) }^{ 2 } \)
= (4 – 1)²
= (3)² = 9

Question 11.
Solution:
\(\left( x+\frac { 1 }{ 4 } \right) =4 \)
Squaring on both sides

Question 12.
Solution:
\(\left( x-\frac { 1 }{ x } \right) =5 \)
(i) Squaring on both sides

Question 13.
Solution:
(i) (x + 1) (x – 1) (x² + 1)
= {(x)² – (1)²} (x² + 1)
= (x² – 1) (x² + 1)
= (x²)² – (1)² = x⁴ – 1 Ans.
(ii) (x – 3) (x + 3) (x² + 9)
= {(x)² – (3)² } (x² + 9)
= (x² – 9) (x² + 9)
= (x²)² – (9)² = x⁴ – 81 Ans.
(iii) (3x – 2y) (3x + 2y) (9x² + 4y²)
= {(3x)² – (2y)²} (9x² + 4y²)
= (9x² – 4y²) (9x² + 4y²)
= (9x²)² – (4y²)²
= 81x⁴ – 16y⁴ Ans.
(iv) (2p + 3) (2p – 3) (4p² + 9)
= {(2p)² – (3)²} (4p² + 9)
= (4p² – 9) (4p² + 9)
= (4p²)² – (9)² = 16p⁴ – 81 Ans.

Question 14.
Solution:
x + y = 12
Squaring both sides,
(x + y)² = (12)²
=> x² + y² + 2xy = 144
=> x² + y² + 2 x 14 = 144
=> x² + y² + 28 = 144
=> x² + y² = 144 – 28 = 116
x² + y² = 116 Ans.

Question 15.
Solution:
x – y = 7
Squaring both sides,
(x – y)² = (7)²
=> x² + y² – 2xy = 49
=> x² + y² – 2 x 9 = 49
=> x² + y² – 18 = 49
=> x² + y² = 49 + 18 = 67
x² + y² = 67 Ans.

Hope given RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6D are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.