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]


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