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Question

If Δ=∣ ∣xabbxaabx∣ ∣ then, a factor of Δ is

A
ab+x
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B
x2(ab)x+a2+b2+ab
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C
x2+(a+b)x+a2+b2ab
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D
a+b+x
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Solution

The correct option is B x2+(a+b)x+a2+b2ab
Δ=∣ ∣xabbxaabx∣ ∣
Δ=x(x2ab)a(bxa2)+b(b2+ax)
Δ=x3+3abx+a3+b3
Δ=x3+3abx+(a+b)33ab(a+b)
Δ=(a+b)3x3+3ab(xab)
Δ=(xab)(x2+(a+b)x)+a2+b2ab

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