RD Sharma class 9 mathematics chapter 5 exercise 5.2 Factorization of Algebraic Expressions is provided here. These exercise solutions are helpful to practice on the factorisation of algebraic expressions expressible as the sum or difference of two cubes. It will also help students score well in the examination. RD Sharma Class 9 chapter 5 solutions are prepared by subject experts using simple and understandable language. Students can download RD Sharma exercise 5.2 by clicking on the link below.
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Access Answers to Maths RD Sharma Class 9 Chapter 5 Factorization of Algebraic Expressions Exercise 5.2 Page number 5.13
Exercise 5.2 Page No: 5.13
Factorize each of the following expressions:
Question 1: p3 + 27
Solution:
p3 + 27
= p3 + 33
[using a3 + b3 = (a + b)(a2 –ab + b2)]= (p + 3)(p² – 3p – 9)
Therefore, p3 + 27 = (p + 3)(p² – 3p – 9)
Question 2: y3 + 125
Solution:
y3 + 125
= y3 + 53
[using a3 + b3 = (a + b)(a2 –ab + b2)]= (y+5)(y2 − 5y + 52)
= (y + 5)(y2 − 5y + 25)
Therefore, y3 + 125 = (y + 5)(y2 − 5y + 25)
Question 3: 1 – 27a3
Solution:
= (1)3 −(3a) 3
[using a3 – b3 = (a – b)(a2 + ab + b2)]= (1− 3a)(12 + 1×3a + (3a) 2)
= (1−3a)(1 + 3a + 9a2)
Therefore, 1−27a3 = (1−3a)(1 + 3a+ 9a2)
Question 4: 8x3y3 + 27a3
Solution:
8x3y3 + 27a3
= (2xy) 3 + (3a) 3
[using a3 + b3 = (a + b)(a2 –ab + b2)]= (2xy +3a)((2xy)2−2xy×3a+(3a) 2)
= (2xy+3a)(4x2y2 −6xya + 9a2)
Question 5: 64a3 − b3
Solution:
64a3 − b3
= (4a)3−b3
[using a3 – b3 = (a – b)(a2 + ab + b2)]= (4a−b)((4a)2 + 4a×b + b2)
=(4a−b)(16a2 +4ab+b2)
Question 6: x3 / 216 – 8y3
Solution:
x3 / 216 – 8y3
Question 7: 10x4 y – 10xy4
Solution:
10x4 y – 10xy4
= 10xy(x3 − y3)
[using a3 – b3 = (a – b)(a2 + ab + b2)]= 10xy (x−y)(x2 + xy + y2)
Therefore, 10x4 y – 10xy4 = 10xy (x−y)(x2 + xy + y2)
Question 8: 54x6 y + 2x3y4
Solution:
54x6 y + 2x3y4
= 2x3y(27x3 +y3)
= 2x3y((3x) 3 + y3)
[using a3 + b3 = (a + b)(a2 – ab + b2)]= 2x3y {(3x+y) ((3x)2−3xy+y2)}
=2x3y(3x+y)(9x2 − 3xy + y2)
Question 9: 32a3 + 108b3
Solution:
32a3 + 108b3
= 4(8a3 + 27b3)
= 4((2a) 3+(3b) 3)
[using a3 + b3 = (a + b)(a2 – ab + b2)]= 4[(2a+3b)((2a)2−2a×3b+(3b) 2)]
= 4(2a+3b)(4a2 − 6ab + 9b2)
Question 10: (a−2b)3 − 512b3
Solution:
(a−2b)3 − 512b3
= (a−2b)3 −(8b) 3
[using a3 – b3 = (a – b)(a2 + ab + b2)]= (a −2b−8b) {(a−2b)2 + (a−2b)8b + (8b) 2}
=(a −10b)(a2 + 4b2 − 4ab + 8ab − 16b2 + 64b2)
=(a−10b)(a2 + 52b2 + 4ab)
Question 11: (a+b)3 − 8(a−b)3
Solution:
(a+b)3 − 8(a−b)3
= (a+b)3 − [2(a−b)]3
= (a+b)3 − [2a−2b] 3
[using p3 – q3 = (p – q)(p2 + pq + q2)]Here p = a+b and q = 2a−2b
= (a+b−(2a−2b))((a+b)2+(a+b)(2a−2b)+(2a−2b) 2)
=(a+b−2a+2b)(a2+b2+2ab+(a+b)(2a−2b)+(2a−2b) 2)
=(a+b−2a+2b)(a2+b2+2ab+2a2−2ab+2ab−2b2+(2a−2b) 2)
=(3b−a)(3a2+2ab−b2+(2a−2b) 2)
=(3b−a)(3a2+2ab−b2+4a2+4b2−8ab)
=(3b−a)(3a2+4a2−b2+4b2−8ab+2ab)
=(3b−a)(7a2+3b2−6ab)
Question 12: (x+2)3 + (x−2) 3
Solution:
(x+2)3 + (x−2) 3
[using p3 + q3 = (p + q)(p2 – pq + q2)]Here p = x + 2 and q = x – 2
= (x+2+x−2)((x+2)2−(x+2)(x−2)+(x−2) 2)
= 2x(x2 +4x+4−(x+2)(x−2)+x2−4x+4)
[ Using : (a+b)(a−b) = a2−b2 ]= 2x(2x2 + 8 − (x2 − 22))
= 2x(2x2 +8 − x2 + 4)
= 2x(x2 + 12)
Access other exercise solutions of Class 9 Maths Chapter 5 Factorization of Algebraic Expressions
RD Sharma Solutions for Class 9 Maths Chapter 5 Exercise 5.2
Class 9 Maths Chapter 5 Factorization of Algebraic Expressions Exercise 5.2 is based on the following topic:
- Factorisation of algebraic expressions expressible as the sum or difference of two cubes
a3 + b3 = (a + b)(a2 + b2 – ab)
a3 – b3 = (a – b)(a2 + b2 + ab)