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Question

If a, b, c are real numbers such that b+cc+aa+bc+aa+bb+ca+bb+cc+a=0, then show that either a+b+c=0 or, a=b=c.

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Solution

Let Δ =b+c c+a a+bc+a a+b b+ca+b b+c c+a =2a+b+c 2a+b+c 2a+b+c c+a a+b b+c a+b b+c c+a Applying R1R1+R2+R3=2a+b+c 1 1 1c+a a+b b+ca+b b+c c+a =2a+b+c 1 0 0c+a b-c b-aa+b c-a c-b Applying C2C2-C1 and C3C3-C1=2a+b+c1b-cb-ac-ac-b=2a+b+cb-cc-b-b-ac-a=-2a+b+ca2+b2+c2-ab-bc-ca=-a+b+c2a2+2b2+2c2-2ab-2bc-2ca=-a+b+ca-b2+b-c2+c-a2But Δ=0 Given-a+b+ca-b2+b-c2+c-a2=0Either a+b+c=0 or a-b2+b-c2+c-a2=0a+b+c=0 or a=b=cHence proved.

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