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Question

If a13+b13+c13=0 then (a+b+c)3=27abc how?

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Solution


Let ay3+by3+cy3=0ABC
Then we know A3+B3+C3=3ABC when (A+B+C=0)
That is, (ay3)3+(by3)3+(cy3)3=3ay3by3cy3
a+b+c=3ay3by3cy3
Cubing both sides,
(a+b+c)3=27abc Proved.

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