# RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions Ex 5.4

## RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions Ex 5.4

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 5.4

Other Exercises

Factorize each of the following expressions:
Question 1.
a3 + 8b3 + 64c3 – 24abc
Solution:
We know that
a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
a3 + 8b3 + 64c3 – 24abc
= (a)3 + (2b)3 + (4c)3 – 3 x a x 2b x 4c
= (a + 2b + 4c) [(a)2 + (2b)2 + (4c)2 -a x 2b – 2b x 4c – 4c x a]
= (a + 2b + 4c) (a2 + 4b2 + 16c2 – 2ab – 8bc – 4ca)

Question 2.
x3 – 8y3 + 27z3 + 18xyz
Solution:
x3 – 8y3 + 27z3 + 18xyz
= (x)3 + (-2y)3 + (3z)3 – 3 x x x (-2y) (3 z)
= (x – y + 3z) (x2 + 4y2 + 9z2 + 2xy + 6yz – 3zx)

Question 3.
27x3 – y3 – z3 – 9xyz          [NCERT]
Solution:
27x3-y3-z3-9xyz
= (3x)3 + (-y)3 + (-z)3 – 3 x 3x x (-y) (-z)
= (3x – y – z) [(3x)2 + (-y)2 + (-z)2 – 3x x (-y) – (-y) (-z)-  (- z x 3x)]
= (3x-y – z) (9x2y2 + z2 + 3xy – yz + 3zx)

Question 4. Solution: Question 5.
8x3 + 27y3 – 216z3 + 108xyz
Solution:
8x3 + 27y3 – 216z3 + 108xyz
= (2x)3 + (3y)3 + (6z)3 – 3 x (2x) (3y) (-6z)
= (2x + 3y – 6z) [(2x)2 + (3y)2 + (-6z)2 – 2x x 3y – 3y x (-6z) – (-6z) x 2x]
= (2x + 3y – 6z) (4x2 + 92 + 36z2 – 6xy + 18yz + 12zx)

Question 6.
125 + 8x3 – 27y3 + 90xy
Solution:
125 + 8X3 – 27y3 + 90xy
= (5)3 + (2x)3 + (-3y)3 – [3 x 5 x 2x x (-3y)]
= (5 + 2x – 3y) [(5)2 + (2x)2 + (-3y)2 – 5 x 2x – 2x (-3y) – (-3y) x 5]
= (5 + 2x – 3y) (25 + 4x2 + 9y2– 10x + 6xy + 15y)

Question 7.
8x3 – 125y3 + 180xy + 216
Solution:
8x3 – 125y3 + 180xy + 216
= (2x)3 + (-5y)3 + (6)3 – 3 x 2x (-5y) x 6
= (2x – 5y + 6) [(2x)2 + (-5y)2 + (6)2 – 2x x (-5y) – (-5y) x 6 – 6 x 2x]
= (2x -5y + 6) (4x2 + 25y2 + 36 + 10xy + 30y – 12x)

Question 8.
Multiply:
(i) x2 +y2 + z2 – xy + xz + yz by x + y – z
(ii) x2 + 4y2 + z2 + 2xy + xz – 2yz by x- 2y-z
(iii) x2 + 4y2 + 2xy – 3x + 6y + 9 by x – 2y + 3
(iv) 9x2 + 25y2 + 15xy + 12x – 20y + 16 by 3x  – 5y + 4
Solution:
(i)  (x2 + y2 + z2 – xy + yz + zx) by (x + y – z)
= x3 +y3 – z3 + 3xyz
(ii) (x2 + 4y2 + z2 + 2xy + xz – 2yz) by (x – 2y – z)
= (x -2y-z) [x2 + (-2y)2 + (-z)2 -x x (- 2y) – (-2y) (z) – (-z) (x)]
= x3 + (-2y)3 + (-z)3 – 3x (-2y) (-z)
= x3 – 8y3 – z3 – 6xyz
(iii) x2 + 4y2 + 2xy – 3x + 6y + 9 by x – 2y + 3
= (x – 2y + 3) (x2 + 4y2 + 9 + 2xy + 6y – 3x)
= (x)3 + (-2y)3 + (3)3 – 3 x x x (-2y) x 3 = x3 – 8y3 + 27 + 18xy
(iv) 9x2 + 25y3 + 15xy + 12x – 20y + 16 by 3x – 5y + 4
= (3x -5y + 4) [(3x)2 + (-5y)2 + (4)2 – 3x x (-5y) (-5y x 4) – (4 x 3x)]
= (3x)3 + (-5y)3 + (4)3 – 3 x 3x (-5y) x 4
= 27x3 – 12573 + 64 + 180xy

Question 9.
(3x – 2y)3 + (2y – 4z)3 + (4z – 3x)3
Solution:
(3x – 2y)3 + (2y – 4z)+   (4z – 3x)3
∵ 3x – 2y + 2y – 4z + 4z – 3x = 0
∴ (3x – 2y)3 + (2y – 4z)3 + (4z – 3x)3
= 3(3x – 2y) (2y – 4z) (4z – 3x)               {∵ x3 + y3 + z3 = 3xyz if x + y + z = 0}

Question 10.
(2x – 3y)3 + (4z – 2x)3  + (3y – 4z)3
Solution:
(2x – 3y)3 + (4z – 2x)3 + (3y – 4z)3
∵  2x – 3y + 4z – 2x + 3y – 4z = 0
∴ (2x – 3y)3 + (4z – 2x)3 + (3y – 4z)3
= (2x – 3y) (4z – 2x) (3y – 4z)                {∵ x3 + y3 + z3 = 3xyz if x + y + z = 0}

Question 11. Solution:  Question 12.
(a – 3b)3 + (3b – c)3 + (c – a)3
Solution:
(a- 3b)3 + (3b – c)3 + (c – a)3
∵ a – 3b + 3b – c + c – a = 0
∴  (a – 3b)3 + (3b – c)3 + (c – a)3
= 3(a – 3b) (3b – c) (c – a)                       {∵ a3 + b3 + c3 = 3abc if a + b + c = 0}

Question 13. Solution: Question 14. Solution: Question 15.
2 $$\sqrt { 2 }$$ a3+ 16$$\sqrt { 2 }$$ b3 + c3 – 12abc
Solution: Question 16.
Find the value of x3 + y– 12xy + 64, when x + y = -4
Solution:
x3 + y– 12xy + 64
x + y = -4
Cubing both sides,
x3 + y3 + 3 xy(x + y) = -64
Substitute the value of (x + y)
⇒ x2 + y2 + 3xy x (-4) = -64
⇒  x3 + y2 – 12xy + 64 = 0

Hope given RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions Ex 5.4 are helpful to complete your math homework.

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