If 2w+1=3i, then the determinant Δ=1111−1−w2w21w2w4 is

w=−12+i32,w3=1, and Δ=3w−3w2

If 2w+1=3i, then the determinant Δ=1111−1−w2w21w2w4 is

w=−12+i32,w3=1, and Δ=3w−3w2

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If $2 w + 1 = \sqrt{3} i$ , then the determinant $Δ = |\begin{array}{ccc} 1 & 1 & 1 \\ 1 & - 1 - w^{2} & w^{2} \\ 1 & w^{2} & w^{4} \end{array}|$ is

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3w²

3w(w – 1)

3w(1 – w)

answer is B.

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$w = \frac{- 1}{2} + \frac{i \sqrt{3}}{2}, w^{3} = 1, and Δ = 3 w - 3 w^{2}$

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If 2w+1=3i, then the determinant Δ=1111−1−w2w21w2w4 is

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If $2 w + 1 = \sqrt{3} i$ , then the determinant $Δ = |\begin{array}{ccc} 1 & 1 & 1 \\ 1 & - 1 - w^{2} & w^{2} \\ 1 & w^{2} & w^{4} \end{array}|$ is