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


Since,


(5a – 7b) + (9c – 5a) + (7b – 9c) = 5a – 7b + 9c – 5a + 7b – 9c = 0


Therefore,


(5a – 7b)3 + (9c – 5a)3 + (7b – 9c)3 = 3 (5a – 7b) (9c – 5a) (7b – 9c)


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