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Question

Let ab+bc+ca=0 then prove that 1a2bc+1b2ac+1c2ab=0.

Solution
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Given (ab+bc+ca)=0 ; ... (1)
Now, from (1) we get,
a2(bc)=a2+ab+ac
or, (a2bc)=a(a+b+c) _ _ _ (2)
similarly, b2ac=b(a+b+c) ... (3)
and c2ab=c(a+b+c) ...(4)
Now,
1a2bc+1b2ac+1c2ab
=1(a+b+c){1a+1b+1c} [using (4),(2) and (3)]
=(ab+bc+ca)(abc)(a+b+c)=0 [using (1)].


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