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(C) Given the determinant $\Delta = \left| \begin{array}{ccc} 1 & \omega^3 & \omega^2 \ \omega^3 & 1 & \omega \ \omega^2 & \omega & 1 \end{array}

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(C) Given the determinant $\Delta = \left| \begin{array}{ccc} 1 & \omega^3 & \omega^2 \ \omega^3 & 1 & \omega \ \omega^2 & \omega & 1 \end{array} ight|$.Since $\omega$ is an imaginary cube root of unity,we know that $\omega^3 = 1$.Substituting $\omega^3 = 1$ into the determinant,we get:$\Delta = \left| \begin{array}{ccc} 1 & 1 & \omega^2 \ 1 & 1 & \omega \ \omega^2 & \omega & 1 \end{array} ight|$.Observe that the first two columns are identical $(C_1 = C_2)$.According to the properties of determinants,if any two columns or rows are identical,the value of the determinant is $0$.Therefore,$\Delta = 0$.

If $\omega$ is one of the imaginary cube roots of unity,then the value of the determinant $\left| \begin{array}{ccc} 1 & \omega^3 & \omega^2 \\ \omega^3 & 1 & \omega \\ \omega^2 & \omega & 1 \end{array} \right|$ is:

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