$\frac{(\sin \frac{\pi}{8} + i \cos \frac{\pi}{8})^8}{(\sin \frac{\pi}{8} - i \cos \frac{\pi}{8})^8}$ is equal to

The sum of the $162^{\text{th}}$ power of the roots of the equation $x^{3}-2x^{2}+2x-1=0$ is

If $1, \omega, \omega^2, \ldots, \omega^{10}$ are the $11^{th}$ roots of unity,then the product of these roots is:

If $\alpha, \beta, \gamma$ are the cube roots of $p$ $(p < 0)$,then for any $x, y$ and $z$,$\frac{x\alpha + y\beta + z\gamma}{x\beta + y\gamma + z\alpha} = $

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

If $Z_{r} = \sin \frac{2 \pi r}{11} - i \cos \frac{2 \pi r}{11}$, then $\sum_{r=0}^{10} Z_{r}$ is equal to