If ϖ is a cube root of unity, then1ω2ω222ω24ω333ω36ω4=

1w2w222w24336w=1(12-12)-2(6w2-6w2)+3(4w-4w)=0

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If $ϖ$ is a cube root of unity, then
$|\begin{matrix} 1 & ω & 2 ω^{2} \\ 2 & 2 ω^{2} & 4 ω^{3} \\ 3 & 3 ω^{3} & 6 ω^{4} \end{matrix}| =$

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Detailed Solution

$|\begin{matrix} 1 & w & 2 w^{2} \\ 2 & 2 w^{2} & 4 \\ 3 & 3 & 6 w \end{matrix}| = 1 (12 - 12) - 2 (6 w^{2} - 6 w^{2}) + 3 (4 w - 4 w) = 0$

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