If $1, \omega, \omega^2$ are the cube roots of unity,then their product is

If $\theta = \frac{\pi}{6}$,then the $10^{th}$ term of the series $1 + (\cos \theta + i \sin \theta) + (\cos \theta + i \sin \theta)^2 + (\cos \theta + i \sin \theta)^3 + \ldots$ is equal to:

If $1, \omega, \omega^2$ are the cube roots of unity,then $\omega^2(1 + \omega)^3 - (1 + \omega^2)\omega = $

$(1+\sqrt{5}+i \sqrt{10-2 \sqrt{5}})^5=$

$(-1+i \sqrt{3})^{60} = ?$

If $\omega$ is a complex cube root of unity and $x = \omega^2 - \omega + 2$,then: