What is the value of $(1-i \sqrt{3})^9$?

If $1, \omega, \omega^2$ are the cube roots of unity,then
$1(2+\frac{1}{\omega})(2+\frac{1}{\omega^2})+2(3+\frac{1}{\omega})(3+\frac{1}{\omega^2})+3(4+\frac{1}{\omega})(4+\frac{1}{\omega^2})+\ldots 10 \text{ terms} =$

If $\alpha, \beta$ are the roots of $1+x+x^2=0$,then the value of $\alpha^4+\beta^4+\alpha^{-4}\beta^{-4}$ is

If $x^{2}+x+1=0$, then the value of $(x+\frac{1}{x})^{4}+(x^{2}+\frac{1}{x^{2}})^{4}+(x^{3}+\frac{1}{x^{3}})^{4}+\dots+(x^{25}+\frac{1}{x^{25}})^{4}$ is:

If $\omega$ is the complex cube root of unity,then the value of $\left(3+5 \omega+3 \omega^2\right)^2+\left(3+3 \omega+5 \omega^2\right)^2$ is:

The product $(1 - \omega + {\omega ^2})(1 - {\omega ^2} + {\omega ^4})(1 - {\omega ^4} + {\omega ^8}) \dots$ up to $2n$ factors is: