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Question

Prove that :

aa+ba+2ba+2baa+ba+ba+2ba=9 a+b b2

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Solution

Let LHS== a a+b a+2ba+2b a a+ba+b a+2b a Δ =3a+3b 3a+3b 3a+3b a+2b a a+ba+b a+2b a Applying R1R1+R2+R3=3 a+b1 1 1 a+2b a a+ba+b a+2b a Taking out 3 a+b common from R1= 3 a+b 0 0 12b -b a+b-b 2b a Applying C1 C1-C2and C2C2-C3 = 3 a+b b2 0 0 1 2 -1 a+b-1 2 a Taking out b common from C1 and C2= 3 a+b b2×3=9a+b b2=RHS

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