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Class 9 RD Sharma Solutions – Chapter 5 Factorisation of Algebraic Expressions- Exercise 5.2 | Set 1
  • Last Updated : 23 Feb, 2021

Factorize each of the following expressions:

Question 1. p3+27

Solution:

⇒ p3+33

⇒(p+3)(p2-3p-9)                                           [ a3+b3=(a+b)(a2-ab+b2) ]

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

Question 2. y3+125

Solution:



⇒ y3+53

⇒ (y+5)(y2-3y+25)                                           [ a3+b3=(a+b)(a2-ab+b2) ]

Therefore, y3+125 = (y+5)(y2-3y+25)

Question 3. 1 – 27a3

Solution:

⇒ (1)3 – (3a)3

⇒ (1-3a)(12+3a+(3a)2)                                [ a3-b3=(a-b)(a2+ab+b2) ]

⇒ (1-3a)(1+3a+9a2)

Therefore, 1 – 27a3 = (1-3a)(1+3a+9a2)



Question 4. 8x3y3 +27a3

Solution:

⇒ (2xy)3 + (3a)3

⇒ (2xy+3a)((2xy)2 -2xy *3a +(3a)2)                [ a3+b3=(a+b)(a2-ab+b2) ]

⇒ (2xy+3a)(4x2y2 -6xya +9a2)

Therefore, 8x3y3+27a3 = (2xy+3a)(4x2y2 -6xya +9a2)

Question 5. 64a3 – b3

Solution:

⇒ (4a)3 – b3

⇒ (4a – b ) ((4a)2 – 4a*b + b2)                                      [ a3-b3=(a-b)(a2+ab+b2) ]

⇒ (4a – b ) (16a2 – 4a*b + b2)

Therefore, 64a3 – b3 = (4a – b ) (16a2 – 4a*b + b2)

Question 6. \frac{x^3}{216}-8y^3

Solution:

(\frac{x}{6})^3-(2y)^3

(\frac{x}6-2y)((\frac{x}{6})^2+\frac{xy}{3}+(2y)^2)               [ a3-b3=(a-b)(a2+ab+b2) ]

(\frac{x}6-2y)(\frac{x^2}{36}+\frac{xy}{3}+4y^2)

Therefore, \frac{x^3}{216}-8y^3 = (\frac{x}6-2y)(\frac{x^2}{36}+\frac{xy}{3}+4y^2)

Question 7. 10x4y – 10xy4

Solution:

⇒ 10xy(x3-y3)

⇒ 10xy(x-y)(x2+xy+y2)                                            [ a3-b3=(a-b)(a2+ab+b2) ]

Therefore, 10x4y – 10xy4 = 10xy(x-y)(x2+xy+y2)

Question 8. 54x6y+2x3y4

Solution:

⇒ 2x3y(27x3+y3)

⇒ 2x3y((3x)3+y3)                                            [ a3+b3=(a+b)(a2-ab+b2) ]

⇒ 2x3y(3x+y)(9x2-3xy+y2)

Therefore, 54x6y+2x3y4 = 2x3y(3x+y)(9x2-3xy+y2)

Question 9. 32a3+108b3

Solution:

⇒ 4(8a3+27b3)

⇒ 4((2a)3+(3b)3)

⇒ 4 [ (2a+3b) ((2a)2 -6ab+(3b)2)]                              [ a3+b3=(a+b)(a2-ab+b2) ]

⇒ 4(2a+3b)(4a2 -6ab +3b2)

Therefore, 32a3+108b3 = 4(2a+3b)(4a2 -6ab +3b2)

Question 10. (a-2b)3-512b3

Solution:

⇒ (a-2b)3 – (8b)3

⇒ (a-2b-8b)((a-2b)2 – (a-2b)8b + (8b)2)         [ a3-b3=(a-b)(a2+ab+b2) ]

⇒ (a – 10b)(a2 +4b2 -2ab +8ab + 16b2 +64b2)

⇒ (a – 10b)(a2 +52b2 +4ab)

Therefore, (a-2b)3-512b3 = (a – 10b)(a2 +52b2 +4ab)

Question 11. (a+b)3 -8(a-b)3

Solution:

⇒ (a+b)3 -[2(a-b)]3

⇒ (a+b)3 – (2a -2b)3

⇒ [a+b-(2a-2b)]((a+b)2+(a+b)(2a-2b)+(2a-2b)2    [a3-b3=(a-b)(a2+ab+b2) ]

⇒ (3b-a)[(a2+b2+2ab) +(2a2-2ab+2ab-2b2)+(4a2+4b2+8ab)]

⇒ (3b-a) [7a2+3b2 -6ab]

Therefore, (a+b)3 -8(a-b)3 = (3b-a) [7a2+3b2 -6ab]

Question 12. (x+2)3(x-2)3

Solution:

⇒ (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)                      [(a+b)(a-b)=a2-b2]

⇒2x(2x2+8-(x2-22))

⇒ 2x(x2 +8+4)

⇒ 2x(x2+12)

Therefore, (x+2)3(x-2)3 = 2x(x2+12)

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