Without finding the cubes, factorise ( x −2 y ) 3 + (2 y −3 z ) 3 + (3 z − x ) 3

Question

Philoid · Accepted Answer

Given that: (x– 2y)³ + (2y – 3z)³ + (3z − x)³

Let a = x – 2y, b= y− 3z and c= 3z − x

∴ a + b + c = x – 2y + 2y – 3z + 3z − x = 0

⟹ (x – 2y)³ + (2y – 3z)³ + (3z − x)³ =

3(x – 2y)(2y – 3z)(z − x)

[∵ a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c²− ab− bc− ca)]

[if a + b + c = 0, a + b³ + c³ = 3abc]

Without finding the cubes, factorise(x−2y)3 + (2y−3z)3 + (3z−x)3