Question

Without finding the cubes, factorise:

(x – 2y)³ + (2y – 3z)³ + (3z – x)³

Answer 1

We know that,

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

Also, if a + b + c = 0, then a³ + b³ + c³ = 3abc

Here, we see that (x – 2y) + (2y – 3z) + (3z – x) = 0

Therefore, (x – 2y)³ + (2y – 3z)³ + (3z – x)³ = 3(x – 2y)(2y – 3z)(3z – x).