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Question

Prove that
∣ ∣a+b+2cabcb+c+2abcac+a+2b∣ ∣=2(a+b+c)3.

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Solution

Lets take the LHS of the given expression
Δ=∣ ∣a+b+2cabcb+c+2abcac+a+2b∣ ∣
Perform C1C1+C2+C3

=∣ ∣ ∣2(a+b+c)ab2(a+b+c)b+c+2ab2(a+b+c)ac+a+2b∣ ∣ ∣

Now, take 2(a+b+c) common from C1
=2(a+b+c)∣ ∣1ab1b+c+2ab1ac+a+2b∣ ∣

Perform R1R1R2 and R2R2R3

=2(a+b+c)∣ ∣ ∣0(a+b+c)00b+c+a(c+a+b)1ac+a+2b∣ ∣ ∣

Now, take (a+b+c) common from R1 and R2
=2(a+b+c)3∣ ∣0100111ac+a+2b∣ ∣

Now, expand the determinants,
=2(a+b+c)3×1(10)
=2(a+b+c)3 Proved




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