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Question

If a, b, c are all non-zero and a + b + c = 0, prove that a2bc+b2ca+c2ab=3.

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Solution

Given: a+b+c=0As we know that,if a+b+c=0, then a3+b3+c3=3abcTherefore, a3+b3+c3=3abcDividing by abc on both sides, we get a3+b3+c3abc=3abcabc a3abc+b3abc+c3abc=3 a2bc+b2ac+c2ab=3Hence, a2bc+b2ac+c2ab=3.


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