If omega is an imaginary cube root of unity, | vedclass

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(D) Given the determinant $D = \left|\begin{array}{ccc}1+\omega & 0 & -\omega \ 1+\omega^{2} & \omega & -\omega^{2} \ \omega+\omega^{2} & \omega & -

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(D) Given the determinant $D = \left|\begin{array}{ccc}1+\omega & 0 & -\omega \ 1+\omega^{2} & \omega & -\omega^{2} \ \omega+\omega^{2} & \omega & -\omega^{2}\end{array}ight|$.Using the property of cube roots of unity, $1+\omega+\omega^{2} = 0$, we have $1+\omega = -\omega^{2}$, $1+\omega^{2} = -\omega$, and $\omega+\omega^{2} = -1$.Substituting these values into the determinant:$D = \left|\begin{array}{ccc}-\omega^{2} & 0 & -\omega \ -\omega & \omega & -\omega^{2} \ -1 & \omega & -\omega^{2}\end{array}ight|$.Perform the operation $R_{2} 	o R_{2} - R_{3}$:$D = \left|\begin{array}{ccc}-\omega^{2} & 0 & -\omega \ -\omega+1 & 0 & 0 \ -1 & \omega & -\omega^{2}\end{array}ight|$.Expanding along the second column:$D = -\omega \left( (-\omega^{2})(0) - (-\omega)(-\omega+1) ight) = -\omega \left( 0 - (\omega^{2} - \omega) ight) = -\omega (-\omega^{2} + \omega) = \omega^{3} - \omega^{2} = 1 - \omega^{2}$.Wait, let us re-evaluate the expansion carefully. Expanding along the second column:$D = -\omega \left( (-\omega^{2})(0) - (-\omega)(-\omega+1) ight) = -\omega (\omega^{2} - \omega) = -\omega^{3} + \omega^{2} = -1 + \omega^{2}$.Actually, looking at the original determinant, $R_2$ and $R_3$ are identical in the second and third columns. If we subtract $R_3$ from $R_2$, the second row becomes $(1+\omega^{2}-(\omega+\omega^{2}), 0, 0) = (1-\omega, 0, 0)$.$D = (1-\omega) \left|\begin{array}{cc}0 & -\omega \ \omega & -\omega^{2}\end{array}ight| = (1-\omega)(0 - (-\omega^{2})) = (1-\omega)(\omega^{2}) = \omega^{2} - \omega^{3} = \omega^{2} - 1$.

If $\omega$ is an imaginary cube root of unity, then the value of the determinant $\left|\begin{array}{ccc}1+\omega & 0 & -\omega \\ 1+\omega^{2} & \omega & -\omega^{2} \\ \omega+\omega^{2} & \omega & -\omega^{2}\end{array}\right|$ is

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