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

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RD Sharma Solutions for Class 9 Chapter 5 Factorisation of Algebraic Expressions Ex 5.4 Download PDF

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) (9x² + y² + 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)³ +   (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 2–√ a³+ 162–√ b³ + c³ – 12abc
Solution:

Question 16.
Find the value of x³ + y³ – 12xy + 64, when x + y = -4
Solution:
x³ + y³ – 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

All Chapter RD Sharma Solutions For Class 9 Maths

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