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

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(B) Given the determinant equation: $\left|\begin{array}{ccc}x+\omega^2 & \omega & 1 \ \omega & \omega^2 & 1+x \ 1 & x+\omega & \omega^2\end{array}\

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(B) Given the determinant equation: $\left|\begin{array}{ccc}x+\omega^2 & \omega & 1 \ \omega & \omega^2 & 1+x \ 1 & x+\omega & \omega^2\end{array}ight|=0$. Applying the column operation $C_1 ightarrow C_1 + C_2 + C_3$: Since $1 + \omega + \omega^2 = 0$, the first column becomes: $C_1 = \begin{bmatrix} x + \omega^2 + \omega + 1 \ \omega + \omega^2 + 1 + x \ 1 + x + \omega + \omega^2 \end{bmatrix} = \begin{bmatrix} x \ x \ x \end{bmatrix}$. Thus, the determinant is $x \left|\begin{array}{ccc} 1 & \omega & 1 \ 1 & \omega^2 & 1+x \ 1 & x+\omega & \omega^2 \end{array}ight| = 0$. This implies $x = 0$ is one of the solutions.

If $\omega$ is an imaginary cube root of unity and $\left|\begin{array}{ccc}x+\omega^2 & \omega & 1 \\ \omega & \omega^2 & 1+x \\ 1 & x+\omega & \omega^2\end{array}\right|=0$, then one of the values of $x$ is

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