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Question

Using properties of determinants, prove the following
∣ ∣ ∣1+a2b22ab2b2ab1a2+b22a2b2a1a2b2∣ ∣ ∣=(1+a2+b2)3.

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Solution

LHS :
∣ ∣ ∣1+a2b22ab2b2ab1a2+b22a2b2a1a2b2∣ ∣ ∣

Applying C1C1bC3,C2C2+aC3, we get,
=∣ ∣ ∣1+a2+b202b01+a2+b22ab(1+a2+b2)a(1+a2+b2)1a2b2∣ ∣ ∣

Taking out (1+a2+b2) common from C1 and C2, we get,
=(1+a2+b2)2∣ ∣102b012aba1a2b2∣ ∣

Applying R3R3bR1+aR2, we get,
=(1+a2+b2)2∣ ∣102b012a001+a2+b2∣ ∣

Expanding along C1, we get,
=(1+a2+b2)2(1+a2+b2)=(1+a2+b2)3 = RHS

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