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Question

If a + b + c = 0, then prove that (b+c)23bc+(c+a)23ca+(a+b)23ab=1

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Solution

Given
a+b+c=0

a+b=c

Cubing on both sides

(a+b)3=(c)3

a3+b3+3(ab)(a+b)=c3

a3+b3+3ab(c)=c3

a3+b33abc=c3

a3+b3+c3=3abc

=(a+b)23ab+(a)23bc+(b)23ac

=(c)23ab+(a)23bc+(b)23ac

=c23ab+a23bc+b23ac

=c3+a3+b33abc

=3abc3abc=1



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