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Question

Prove that :

1a2+bca31b2+cab31c2+abc3=-a-b b-c c-a a2+b2+c2

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Solution

Let LHS =Δ=1 a2+bc a31 b2+ca b31 c2+ab c3Δ =0 a2+bc-b2+ca a3-b30 b2+ca-c2+ab b3-c31 c2+ab c Applying R1R1-R2 and R2R2- R3

=0 a2-b2-ca+ bc a3-b30 b2-c2-ab+ ca b3-c31 c2+ab c3=0 a-b a+b-c a-ba2+ab+b20 b-cb+ c-a b-cb2+bc+a21 c2+ab c3

=a-bb-c0 a+b-c a2+ab+b20 b+c-a b2 +bc+c21 c2+ab c3 Taking out a-b common from R1 and b-c from R2= a-bb-c0 a+b-c a2+ab+b20 b+c-a- a+b-c b2 +bc+c2-a2+ab+b21 c2+ab c3 Applying R2R2-R1

=a-bb-c0 a+b-c a2+ab+b20 2 c-a bc-a+c2-a21 c2+ab c3=a-bb-cc-a 0 a+b-c a2+ab+b20 2 a+b+c1 c2+ab c3=a-bb-cc-a×1× a+b-c a2+ab+b2 2 a+b+c Expanding along C1


=a-bb-cc-a× a+b2-c2 - 2a2+2ab+2b2 =a-bb-cc-aa+b2-c2 -a+b2-a2+b2=-a-bb-cc-aa2+b2+c2=RHS

Hence proved.

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