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

The square of either of the two imaginary cube roots of unity is:

If $x = a, y = b\omega, z = c\omega^2$,where $\omega$ is a complex cube root of unity,then $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = $

$(\sin \theta + i\cos \theta )^n$ is equal to

If $\omega$ is an imaginary cube root of unity,then the value of $(1-\omega+\omega^{2}) \cdot(1-\omega^{2}+\omega^{4}) \cdot(1-\omega^{4}+\omega^{8}) \cdot \ldots$ ($2n$ factors) is

${\left( \frac{\sqrt{3} + i}{2} \right)^6} + {\left( \frac{i - \sqrt{3}}{2} \right)^6}$ is equal to