If $1,\omega ,{\omega }^2$ the cube roots of unity, then the value of

$\left|\begin{array}{ccc}1& {\omega }^n& {\omega }^{2n}\\ {\omega }^{2n}& 1& {\omega }^n\\ {\omega }^n& {\omega }^{2n}& 1\end{array}\right|$ is equal to 

Let $1,\omega ,{\omega }^2$ be the cube roots of unity. 

$\because {\omega }^3=1,1+\omega +{\omega }^2=0$

Let $A=\left|\begin{array}{ccc}1& {\omega }^n& {\omega }^{2n}\\ {\omega }^{2n}& 1& {\omega }^n\\ {\omega }^n& {\omega }^{2n}& 1\end{array}\right|$

$\begin{array}{l}=1\left(1-{\omega }^{3n}\right)-{\omega }^n\left({\omega }^{2n}-{\omega }^{2n}\right)+{\omega }^{2n}\left({\omega }^{4n}-{\omega }^n\right)\\ =1-{\omega }^{3n}+{\omega }^{6n}-{\omega }^{3n}\\ =1-2{\left({\omega }^3\right)}^n+{\left[{\omega }^3\right]}^{2n}\\ =1-2+1\left(\because {\omega }^3=1\right)\\ =0\end{array}$