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Question

∣ ∣ ∣b2+c2a2a2b2c2+a2b2c2c2a2+b2∣ ∣ ∣ =

A
a2b2c2
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B
4abc
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C
4a2b2c2
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D
2a2b2c2
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Solution

The correct option is D 4a2b2c2
Performing R1R1R2R3
∣ ∣ ∣02c22b2b2c2+a2b2c2c2a2+b2∣ ∣ ∣
(2)∣ ∣ ∣0c2b2b2c2+a2b2c2c2a2+b2∣ ∣ ∣
Performing R3R3R1
(2)∣ ∣ ∣0c2b2b2c2+a2b2c20a2∣ ∣ ∣
Performing C3C3C1
(2)∣ ∣ ∣0c2b2b2c2+a20c20a2c2∣ ∣ ∣
(2)[(c2)(b2a2b2c2)+b2(c4c2a2)]
=4a2b2c2

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