RD Sharma Solutions for Class 9 Mathematics Chapter 5 Exercise 5.2 Factorization of Algebraic Expressions are provided here. These exercise solutions are helpful to practise the factorization of algebraic expressions expressible as the sum or difference of two cubes. It will also help students score well in the examination. RD Sharma Solutions Class 9 Chapter 5 is prepared by subject experts using simple and understandable language. Students can download the RD Sharma Exercise 5.2 by clicking on the link below.
RD Sharma Solutions for Class 9 Maths Chapter 5 Factorization of Algebraic Expressions Exercise 5.2
Access Answers to Maths RD Sharma Solutions for 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)
RD Sharma Solutions for Class 9 Maths Chapter 5 Factorization of Algebraic Expressions Exercise 5.2
RD Sharma Solutions for Class 9 Maths Chapter 5 Factorization of Algebraic Expressions Exercise 5.2 are based on the following topic:
- Factorization 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)
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