RD Sharma Class 9 Solutions Chapter 5 Exercise 5.2 (Updated for 2024)

RD Sharma Class 9 Solutions Chapter 5 Exercise 5.2: You must clear your basic doubts in Maths so as to understand the subject from the core. These exercise solutions are helpful to practice the factorization of algebraic expressions expressible as the sum or difference of two cubes. For this purpose you can use RD Sharma Solutions Class 9 Maths can be very useful. All the solutions of RD Sharma Class 9 Solutions Chapter 5 Exercise 5.2 are designed by subject matter experts which are very reliable. To know more, read the whole blog. 

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RD Sharma Class 9 Solutions Chapter 5 Exercise 5.2

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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)

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