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