Question

Factorise: (2a – b )3 + (b – 2c )3 + 8 (c – a)3​

Answer 1

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

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)