Question

factorise (2a-b)^3+(b-2c) ^3+8(c-a) ^3

Answer 1

since ,when a+b+c=0
a^3+b^3+c^3=3abc

(2a-b)^3+(b-2c) ^3+8(c-a) ^3
=3(2a-b)(b-2c)(2c-2a)

Answer 2

${\mathfrak{\red{\underline{\underline{Solution:-}}}}}$

The given equation is,

(2a - b)³ + (b - 2c)³ + 8(c - a)³

= (2a - b)³ + (b - 2c)³ + (2c - 2a)³

Let x = (2a - b) ; y = (b - 2c) ; z = (2c - 2a)

Now,

x + y+ z = (2a - b) + (b - 2c) + (2c - 2a) = 0

Since, x + y + z = 0, then,

x³ + y³ + z³ = 3xyz

⇒ (2a - b)³ + (b - 2c)³ + (2c - 2a)³ = 3(2a - b) (b - 2c) (2c - 2a)

⇒ (2a - b)³ + (b - 2c)³ + 8(c - a)³ = 6(2a - b) (b - 2c) (c - a)