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If $ω$ is a complex cube root of unity, then a root of the equation $|\begin{array}{ccc} x + 1 & ω & ω^{2} \\ ω & x + ω^{2} & 1 \\ ω^{2} & 1 & x + ω \end{array}| = 0$ is

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x=1

x=ω

x=ω2

x=0

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detailed solution

Correct option is D

Let us denote the given determinant by ∆.Applying C1→C1+C2+C3, we getΔ=x+1+ω+ω2ωω2x+1+ω+ω2x+ω21x+1+ω+ω21x+ω=xωω2xx+ω21x1x+ωClearly ∆=0 for x=0.

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If ω is a complex cube root of unity, then a root of the equation x+1ωω2ωx+ω21ω21x+ω=0 is

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If $ω$ is a complex cube root of unity, then a root of the equation $|\begin{array}{ccc} x + 1 & ω & ω^{2} \\ ω & x + ω^{2} & 1 \\ ω^{2} & 1 & x + ω \end{array}| = 0$ is