If ω is a non-real cube root of unity, then what is a root of the...

Applying the column operation $C_1 \rightarrow C_1 + C_2 + C_3$, the determinant becomes:

\begin{vmatrix} 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{vmatrix} = 0$<br><br>Since 1 + \omega + \omega^2 = 0$ for a non-real cube root of unity, the equation simplifies to:

\begin{vmatrix} x & \omega & \omega^2 \\ x & x+\omega^2 & 1 \\ x & 1 & x+\omega \end{vmatrix} = 0$<br><br>Taking$x$common from the first column:<br><br>$x \begin{vmatrix} 1 & \omega & \omega^2 \\ 1 & x+\omega^2 & 1 \\ 1 & 1 & x+\omega \end{vmatrix} = 0$<br><br>This implies$x = 0$is a root of the equation.<br><br>Answer:$x = 0