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प्रश्न
Without finding the cubes, factorise:
(x – 2y)3 + (2y – 3z)3 + (3z – x)3
उत्तर
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).
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