Let omega = -frac{1}{2} + i frac{sqrt{3}}{2}, | vedclass

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(C) Given $\omega = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$,we know $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$. Since $1 + \omega + \omega^2 = 0$,we

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$3 \omega(\omega-1)$

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(C) Given $\omega = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$,we know $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$. Since $1 + \omega + \omega^2 = 0$,we have $-1 - \omega^2 = \omega$. Also,$\omega^4 = \omega^3 \cdot \omega = 1 \cdot \omega = \omega$. Substituting these into the determinant: $\Delta = \left| \begin{array}{ccc} 1 & 1 & 1 \ 1 & \omega & \omega^2 \ 1 & \omega^2 & \omega \end{array} ight|$ Expanding along the first row: $\Delta = 1(\omega^2 - \omega^4) - 1(\omega - \omega^2) + 1(\omega^2 - \omega)$ $\Delta = 1(\omega^2 - \omega) - 1(\omega - \omega^2) + 1(\omega^2 - \omega)$ $\Delta = \omega^2 - \omega - \omega + \omega^2 + \omega^2 - \omega$ $\Delta = 3\omega^2 - 3\omega$ $\Delta = 3\omega(\omega - 1)$

Let $\omega = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$,where $i = \sqrt{-1}$. Then the value of $\left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & -1-\omega^2 & \omega^2 \\ 1 & \omega^2 & \omega^4 \end{array} \right|$ is

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