Skip to content
Related Articles

Related Articles

Improve Article

Class 9 RD Sharma Solutions – Chapter 5 Factorisation of Algebraic Expressions- Exercise 5.2 | Set 1

  • Last Updated : 23 Feb, 2021
Geek Week

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)

Attention reader! All those who say programming isn’t for kids, just haven’t met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12. 

The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.




My Personal Notes arrow_drop_up
Recommended Articles
Page :