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Question

Prove that :

a+b+2cabcb+c+2abcac+a+2b=2 a+b+c3

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Solution

Let LHS= Δ=a+b+2c a b c b+c+2a b c a c+a +2b = 2a+2b+2c a b 2a+2b+2c b+c+2a b 2a+2b+2c a c+a +2b Applying C1 C1+C2+C 3 = 2 a+b+c 1 a b 1 b+c+2a b1 a c+a +2b Taking out 2(a+b+c) common from C1 =2 a+b+c 1 a b 0 b+c+a 00 -b-c-a c+a +b Applying R2 R2-R1 and R2R2 -R3=2a+b+ca+b+ca+b+c1 a b 0 1 00 -1 1 Taking out (a+b+c) common from R2 and R3=2a+b+c3 11-0 Expanding along C1=2a+b+c3 =RHS

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