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(D) Let $\Delta = \left| \begin{array}{ccc} x + 1 & \omega & \omega^2 \ \omega & x + \omega^2 & 1 \ \omega^2 & 1 & x + \omega \end{array} ight|$.A

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$x^3$

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(D) Let $\Delta = \left| \begin{array}{ccc} x + 1 & \omega & \omega^2 \ \omega & x + \omega^2 & 1 \ \omega^2 & 1 & x + \omega \end{array} ight|$.Applying the column operation $C_1 	o C_1 + C_2 + C_3$,we get:$\Delta = \left| \begin{array}{ccc} x + 1 + \omega + \omega^2 & \omega & \omega^2 \ x + 1 + \omega + \omega^2 & x + \omega^2 & 1 \ x + 1 + \omega + \omega^2 & 1 & x + \omega \end{array} ight|$.Since $1 + \omega + \omega^2 = 0$,the first column becomes $x$:$\Delta = x \left| \begin{array}{ccc} 1 & \omega & \omega^2 \ 1 & x + \omega^2 & 1 \ 1 & 1 & x + \omega \end{array} ight|$.Expanding along the first column:$\Delta = x [ 1((x + \omega^2)(x + \omega) - 1) - 1(\omega(x + \omega) - \omega^2) + 1(\omega - \omega^2(x + \omega^2)) ]$.Simplifying the terms inside:$= x [ (x^2 + \omega x + \omega^2 x + \omega^3 - 1) - (\omega x + \omega^2 - \omega^2) + (\omega - \omega^2 x - \omega^4) ]$.Using $\omega^3 = 1$ and $\omega^4 = \omega$:$= x [ x^2 + \omega x + \omega^2 x + 1 - 1 - \omega x + \omega - \omega^2 x - \omega ] = x [ x^2 ] = x^3$.

If $\omega$ is a cube root of unity,then $\left| \begin{array}{ccc} x + 1 & \omega & \omega^2 \\ \omega & x + \omega^2 & 1 \\ \omega^2 & 1 & x + \omega \end{array} \right| = $

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