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Question

Prove that ∣ ∣ ∣a2+1abacabb2+1bccacbc2+1∣ ∣ ∣=1+a2+b2+c2

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Solution

Let , L.H.S.

Taking out common a,b,c

=abc∣ ∣ ∣ ∣a+1abcab+1bcabc+1c∣ ∣ ∣ ∣

multiplying a,b,c in c1,c2,c3 respectively

=∣ ∣ ∣a2+1b2c2a2b2+1c2a2b2c2+1∣ ∣ ∣

Applying c1c1+c2+c3

=∣ ∣ ∣1+a2+b2+c2b2c21+a2+b2+c2b2+1c21+a2+b2+c2b2c2+1∣ ∣ ∣

Taking out common 1+a2+b2+c2 from c1

=(1+a2+b2+c2)∣ ∣ ∣1b2c21b2+1c21b2c2+1∣ ∣ ∣

Applying
R2R2R1
R3R3R1

=(1+a2+b2+c2)∣ ∣1b2c2010001∣ ∣

Expanding along c1

=1+a2+b2+c2

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