Answer :

We know that,


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


Using this formula, we get


Putting (a – b) = x, (b – c) = y and (c – a) = z, we get


(a - b)3 + (b - c)3 + (c - a)3= x3 + y3 + z3


Where (x + y + z) = (a – b) + (b – c) + (c – a) = 0


= 3xyz [Since, (x + y + z) = 0 so (x3 + y3 + z3) = 3xyz]


= 3 (a – b) (b – c) (c – a)


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