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Question

Prove the following :
∣ ∣ ∣b2+c2abacabc2+a2bccacba2+b2∣ ∣ ∣
=∣ ∣ ∣b2+c2a2a2b2c2+a2b2c2c2a2+b2∣ ∣ ∣=4a2b2c2

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Solution

Take out a, b, c from C1, C2 and C3 respectively.
=abc∣ ∣aca+ca+bbabb+cc∣ ∣
Apply C1+(C2C3) and take 2b common from C1
=abc.2b∣ ∣0ca+c1ba1b+cc∣ ∣
Apply R2R3
=2ab2c∣ ∣0ca+c0cac1b+cc∣ ∣
=2ab2c{c(ac)+c(a+c)}
=2ab2c(2ac)=4a2b2c2

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