If $\omega$ is one of the imaginary cube roots of unity, show that the square of
$\begin{vmatrix}1&\omega&\omega^2&\omega^3\\
\omega&\omega^2&\omega^3&1\\
\omega^2&\omega^3&1&\omega\\
\omega^3&1&\omega&\omega^2\end{vmatrix} = \begin{vmatrix}1&1&-2&1\\1&1&1&-2\\-2&1&1&1\\1&-2&1&1\end{vmatrix}$
Hence, show that the value of the determinant on the left is $3√-3$ .
What I've tried :
I'd multiplied the determinant on left 2 times and I got the following determinant and I don't know how to get the desired determinant from here $\begin{vmatrix}1&1&\omega&1\\ 1&1&1&-2\\ -2&1&1&1\\ 1&-2\omega&1&-2\omega\end{vmatrix}$