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Question

Prove that :

a2bcac+c2a2+abb2acabb2+bcc2=4a2b2c2

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Solution

Let LHS =Δ=a2 bc ac+c2a2+ab b2 acab b2+bc c2Δ=abc a c a+ca+b b ab b+c c Taking out a, b and c common from C1, C2 and C3

=abc a c 0a+b b -2bb b+c -2b Applying C3C3-C2-C1 =abc-2b a c 0a+b b 1b b+c 1 Taking (-2b) common from C3=abc-2b a c 0a -c 0b b+c 1 Applying R2R2-R3 =abc-2b×1 a c a -c Expanding along C3=abc-2b-2ac=4a2b2c2=RHS

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