Show :
`omega^4 + 2 omega^6 + omega^8 = 1`

Show that \begin{vmatrix}1& \omega & (\omega)^2\\ \omega &1 & (\omega)^2\\ \omega & (\omega)^2 & 1 \end{vmatrix} =0

If omega is a complex cube roots of unity, then find the value of the (1+ omega)(1+ omega^(2))(1+ omega^(4)) (1+ omega^(8)) … to 2n factors.

(2 + 5 omega + 2 omega ^ (2)) ^ (6) = (2 + 2 omega + 5 omega ^ (2)) ^ (6) = 729

If 1, omega, omega^2 be the three cube roots of 1, then show that: (1+omega)(1+omega^2)(1+omega^4)(1+omega^8)=1