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Question

If α and β are non-real numbers satisfying x3−1=0, then the value of ∣∣ ∣∣λ+1αβαλ+β1β1λ+α∣∣ ∣∣ is?

A
0
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B
λ3
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C
λ3+1
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D
λ31
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Solution

The correct option is B λ3It is given that α and β are the non-real roots of the equation x3−1=0.We have, x3−1=0.⇒x3=1⇒x=1,ω,ω2Hence, α=ω and β=ω2Now, ∣∣ ∣∣λ+1αβαλ+β1β1λ+α∣∣ ∣∣=∣∣ ∣∣λ+1+α+βλ+1+α+βλ+α+β+1αλ+β1β1λ+α∣∣ ∣∣ [∵R1→R1+R2+R3]=(λ+1+α+β)∣∣ ∣∣111αλ+β1β1λ+α∣∣ ∣∣ [∵C1→C1−C2,C2→C2−C3]=(λ+1+α+β)∣∣ ∣∣001α−λ−βλ+β−11β−11−λ−αλ+α∣∣ ∣∣=(λ+1+α+β)⋅1[(α−λ−β)(1−λ−α)−(β−1)(λ+β−1)]=(λ+1+α+β)(α−α2+λ2+αβ−β2+β−1)=(λ+1+ω+ω2)(ω−ω2+λ2+ω3−ω4+ω2−1) (putting α=ω and β=ω2)=λ(ω−ω2+λ2+1−ω+ω2−1) ..... [∵ω3=1⟹1+ω+ω2=0]=λ3.

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