<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] You typed $\omega=\frac{1\pm\sqrt3\,i}{2}$ but I think you mean
$$\omega=\frac{-1+\sqrt3\,i}{2}\ .$$
In that case, subtracting a suitable multiple of row 1 from every other row (and not bothering to calculate irrelevant entries) gives
$$\eqalign{\det\pmatrix{1&\omega&\omega^2&1\cr \omega&\omega^2&1&1\cr \omega^2&1&1&\omega\cr 1&1&\omega&\omega^2\cr}
  &=\det\pmatrix{1&\omega&\omega^2&1\cr 0&0&0&1-\omega\cr 0&0&1-\omega&\cdot\cr 0&1-\omega&\cdot&\cdot\cr}\cr
  &=-(1-\omega)^3\cr &=3(\omega-\omega^2)\cr &=3\sqrt{-3}\ .\cr}$$
</s>