If $(\sqrt{3}+i)^8-(\sqrt{3}-i)^8=\alpha+i \beta$,then $\alpha-\frac{\sqrt{3}}{2} \beta=$

If $\alpha$ is a non-real root of $x^6=1$,then $\frac{\alpha^5+\alpha^3+\alpha+1}{\alpha^2+1}$ is equal to

The roots of the equation $x^3-3x^2+3x+7=0$ are $\alpha, \beta, \gamma$ and $\omega, \omega^2$ are complex cube roots of unity. If the terms containing $x^2$ and $x$ are missing in the transformed equation when each one of these roots is decreased by $h$,then $\frac{\alpha-h}{\beta-h}+\frac{\beta-h}{\gamma-h}+\frac{\gamma-h}{\alpha-h}=$

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

If $\alpha, \beta$ are the roots of the equation $x^2-2x+4=0$ and for any $n \in N, \alpha^n+\beta^n=k \cos \frac{n \pi}{3}$, then $k=$

If $\omega$ is the complex cube root of unity and $\left(\frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2}\right)^k+\left(\frac{a+b \omega+c \omega^2}{b+a \omega^2+c \omega}\right)^l=2$,then $2k+l$ is always