<s>[INST] calculate one determinant to show 【National College Entrance Exam, old China】
$  \omega  =\frac{1 \pm \sqrt{3} i }  {2}$
Please calculate and show:
$  \begin{vmatrix} 1 &\omega& \omega^2 & 1\\      \omega& \omega^2 & 1 &1\\      \omega^2& 1& 1 & \omega   \\          1& 1&        \omega&  \omega^2\\          \end{vmatrix}    =3\sqrt{-3}   $
Here's what I have done until now
$$
D=\begin{vmatrix}
1 & \omega & {{\omega }^{2}} & 1 \\
\omega & {{\omega }^{2}} & 1 & 1 \\
{{\omega }^{2}} & 1 & 1 & \omega \\
1 & 1 & \omega & {{\omega }^{2}} \\
\end{vmatrix} \\= \begin{vmatrix}
1 & \omega & {{\omega }^{2}} & 1+1+\omega +{{\omega }^{2}} \\
\omega & {{\omega }^{2}} & 1 & 1+1+\omega +{{\omega }^{2}} \\
{{\omega }^{2}} & 1 & 1 & 1+1+\omega +{{\omega }^{2}} \\
1 & 1 & \omega & 1+1+\omega +{{\omega }^{2}} \\
\end{vmatrix} \\= \begin{vmatrix}
1 & \omega & {{\omega }^{2}} & 1 \\
\omega & {{\omega }^{2}} & 1 & 1 \\
{{\omega }^{2}} & 1 & 1 & 1 \\
1 & 1 & \omega & 1 \\
\end{vmatrix} \\= \begin{vmatrix}
1 & \omega & {{\omega }^{2}} & 1 \\
\omega -1 & {{\omega }^{2}}-\omega & 1-{{\omega }^{2}} & 0 \\
{{\omega }^{2}}-1 & 1-\omega & 1-{{\omega }^{2}} & 0 \\
0 & 1-\omega & \omega -{{\omega }^{2}} & 0 \\
\end{vmatrix}
$$
 [/INST] One approach is to note that this matrix (which I will refer to as $A$) is a circulant matrix. As is explained on the page, we can thereby deduce that the eigenvalues of $A$ are of the form
$$
\lambda_j = 1 + i^j + \omega^2 i^{2j} + \omega i^{3j}, \quad i = \sqrt{-1}, \quad j = 0,1,2,3.
$$
So, we find that the eigenvalues are
$$
\lambda_0 = 1 + (1 + \omega + \omega^2) = 1\\
\lambda_1 = 1 + i - \omega^2 - i\omega = (1 + 1/2 + \sqrt{3}/2) +i(1 + \sqrt{3}/2 + 1/2)
\\
\lambda_2 = 1  -1 + \omega^2 - \omega = -i\sqrt{3}\\
\lambda_3 = 1  - i - \omega^2 + i\omega = (1 + 1/2 - \sqrt{3}/2) - i(1 - \sqrt{3}/2 + 1/2).
$$
It now suffices to compute the product of these eigenvalues.
</s>