Answer :

We have,


a3 (b – c)3 + b3 (c – a)3 + c3 (a – b)3


We know that,


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


Also,


If, (x + y + z) = 0


Then (x3 + y3 + z3) = 3xyz


= [a (b – c)]3 + [b (c – a)]3 + [c (a – b)]3


Since,


a (b – c) + b (c – a) + c (a – b) = ab – ac + bc – ba + ca – bc = 0


So,


a3 (b – c)3 + b3 (c – a)3 + c3 (a – b)3


= 3a (b – c) b (c – a) c (a – b)


= 3abc (a – b) (b – c) (c – a)


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