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Q.

If Δ1=1a2a31b2b31c2c3 and Δ2=bcb+c1cac+a1aba+b1 Then Δ1Δ2=

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a

ab+bc+ca

b

abc

c

2(ab+bc+ca)

d

(a+b+c)2

answer is A.

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Detailed Solution

Δ1=(ab+bc+ca)(ab)(bc)(ca)Δ2=(ab)(bc)(ca)Δ1Δ2=ab+bc+ca

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