Q.

If ω≠ 1 is a cube root of unity and Δ=x+ω2ω1ωω21+x1x+ωω2=0, then value  of x is

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a

0

b

1

c

-1

d

none of these

answer is A.

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Detailed Solution

Applying C1→C1+C2+C3 , we get Δ=x+ω2+ω+1ω1ω+ω2+1+xω21+x1+x+ω+ω2x+ωω2=xω1xω21+xxx+ωω2 using 1 +ω+ω2=0∆is clearly equal to 0 for x = 0.
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If ω≠ 1 is a cube root of unity and Δ=x+ω2ω1ωω21+x1x+ωω2=0, then value  of x is