Question:
Without finding the cubes, factorise (x- 2y)3 + (2y – 3z)3 + (3z – x)3.
Solution:
We know that,
a3 + b3 + c3 – 3 abc = (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).
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