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Question

The value of ∣∣ ∣ ∣∣1+a2−b22ab−2b2ab1−a2+b22a2b−2a1−a2−b2∣∣ ∣ ∣∣ is equal to

A
(1+a2+b2)3
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B
(1+a2+b2)2
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C
(1a2+b2)2
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D
(a2b21)3
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Solution

The correct option is A (1+a2+b2)3
The value of ∣ ∣ ∣1+a2b22ab2b2ab1a2+b22a2b2a1a2b2∣ ∣ ∣ is

C1C1bC3,C2C2+aC3

=∣ ∣ ∣1+a2+b202b01+a2+b22ab(1+a2+b2)a(1+b2+a2)1a2b2∣ ∣ ∣

=(1+a2+b2)2∣ ∣102b012aba1a2b2∣ ∣

=(1+a2+b2)2{(1a2b2+2a2)+2b2}
=(1+a2+b2)3

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