RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions Ex 5.3

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions Ex 5.3

Other Exercises

Factorize:
Question 1.
64a3 + 125b3 + 240a2b + 300ab2
Solution:
64a3 + 125b3 + 240a2b + 300ab2
= (4a)3 + (5b)3 + 3 x (4a)2 x 5b + 3(4a) + (5b)2
= (4a + 5b)3
= (4a + 5b) (4a + 5b) (4a + 5b)

Question 2.
125x3 – 27y3 – 225x2y + 135xy2
Solution:
125x3 – 27y3 – 225x2y + 135xy2
= (5x)3 – (3y)3 – 3 x (5x)2 x (3y) + 3- x 5x x (3y)2
= (5x – 3y)3
= (5x – 3y) (5x – 3y) (5x – 3y)

Question 3.
RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions Ex 5.3 Q3.1
Solution:
RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions Ex 5.3 Q3.2

Question 4.
8x3 + 27y3 + 36x2y + 54xy2
Solution:
8x3 + 27y3 + 16x2y + 54xy2
= (2x)3 + (3y)3 + 3 x (2x)2 x 3y  +  3 x 2x x (3y)2
= (2x + 3y)3
= (2x + 3y) (2x + 3y) (2x + 3y)

Question 5.
a3 – 3a2b + 3ab2 – b3 + 8
Solution:
a3 – 3a2b + 3ab2 – b3 + 8
= (a – b)3 + (2)3
= (a – b + 2) [(a -b)2– (a – b) x 2 + (2)2]
= (a- b + 2) (a2 + b2 -2ab – 2a + 2b + 4)

Question 6.
x3 + 8y3 + 6x2y + 12xy2
Solution:
x3 + 8y3 + 6x2y + 12xy2
= (x)3 + (2y)3 + 3 x x2x 2y + 3 x x x (2y)2
= (x + 2y)3
= (x + 2y) (x + 2y) (x + 2y)

Question 7.
8x3 + y3 + 12x2y + 6xy2
Solution:
8x3 + y3 + 12x2y + 6xy2
= (2x)3 + (y)3 + 3 x (2x)2 x y + 3 x 2x x y2
= (2x + y)3
= (2x + y) (2x + y) (2x + y)

Question 8.
8a3 + 27b3 + 36a2b + 54ab2
Solution:
8a3 + 27b3 + 16a2b + 54ab2
= (2a)3 + (3b)3 + 3 x (2a)2 x 3b + 3 x 2a x (3b)2
= (2a + 3b)3
= (2a + 3b) (2a + 3b) (2a + 3b)

Question 9.
8a3 – 27b3 – 36a2b + 54ab2
Solution:
8a3 – 27b3 – 36a2b + 54ab2
= (2a)3 – (3b)3 – 3 x (2a)2 x 3b + 3 x 2a x (3b)2
= (2a – 3b))3
= (2a – 3b) (2a – 3b) (2a – 3b)

Question 10.
x3 – 12x(x – 4) – 64
Solution:
x3 – 12x(x – 4) – 64
= x3 – 12x2 + 48x – 64
= (x)3 – 3 x x2 x 4 + 3 x x x (4)2– (4)3
= (x – 4)3
= (x – 4) (x – 4) (x – 4)

Question 11.
a3x3 – 3a2bx2 + 3ab2x – b3
Solution:
a3x3 – 3a2bx2 + 3ab2x – b3
= (ax)3 – 3 x (ax)2 x  b + 3 x ax x (b)2– (b)3
= (ax – b)3
= (ax – b) (ax – b) (ax – b)

 

Hope given RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions Ex 5.3 are helpful to complete your math homework.

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