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Question

If ω1 is a cube root of unity and ∣ ∣ ∣x+ω2ω1ωω21+x1x+ωω2∣ ∣ ∣=0, then value of x is


A

0

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B

1

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C

-1

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D

none of these

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Solution

The correct option is A

0


Let Δ=∣ ∣ ∣x+ω2ω1ωω21+x1x+ωω2∣ ∣ ∣
Applying C1 C1+C2+C3 we get
Δ=∣ ∣ ∣x+ω2+ω+1ω1ω+ω2+1+xω21+x1+x+ω+ω2x+ωω2∣ ∣ ∣=∣ ∣ ∣xω1xω21+xxx+ωω2∣ ∣ ∣ (1+ω+ω2=0)
Δ is clearly equal to 0 for x = 0.


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