RD Sharma Solutions for Class 9 Maths Chapter 5 Factorization of Algebraic Expressions Exercise 5.2

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 Solutions Class 9 chapter 5 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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Exercise 5.2 Page No: 5.13

Factorize each of the following expressions:

Question 1: p³ + 27

Solution:

p³ + 27

= p³ + 3³

= (p + 3)(p² – 3p – 9)

Therefore, p³ + 27 = (p + 3)(p² – 3p – 9)

Question 2: y³ + 125

Solution:

y³ + 125

= y³ + 5³

= (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

= (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

= (2xy +3a)((2xy)²−2xy×3a+(3a) ²)

= (2xy+3a)(4x²y² −6xya + 9a²)

Question 5: 64a³ − b³

Solution:

64a³ − b³

= (4a)³−b³

= (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)

= 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)

= 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)

= 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

= (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

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

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)

= 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 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)