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Question

Using properties of determinants it can be proved
∣ ∣b+caabc+abcca+b∣ ∣=4abc

A
True
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B
False
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Solution

The correct option is A True
∣ ∣b+caabc+abcca+b∣ ∣
=∣ ∣02c2bbc+abcca+b∣ ∣[R1=R1(R2+R3)]
expanding w.r.t. row 1, we get,
=0+2c(b(a+b)bc))2b(bcc(c+a))=2abc+2b2c2bc22b2c+2bc2+2abc=4abc

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