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

The general value of the real angle $\theta$, which satisfies the equation $(\cos \theta + i \sin \theta)(\cos 2\theta + i \sin 2\theta) \dots (\cos n\theta + i \sin n\theta) = 1$ is given by (assuming $k$ is an integer):

Let $A_r = \left(x+\frac{1}{x}\right)^3 \cdot \left(x^2+\frac{1}{x^2}\right)^3 \cdot \left(x^3+\frac{1}{x^3}\right)^3 \cdots \left(x^r+\frac{1}{x^r}\right)^3$. If $x^2+x+1=0$,then $\frac{1}{A_3}+\frac{1}{A_6}+\frac{1}{A_9}+\frac{1}{A_{12}}+\cdots \infty =$

If $\sqrt{x}+\frac{1}{\sqrt{x}}=2 \cos \theta$,then $x^6+x^{-6}=$

If $\omega$ represents a cube root of unity and $\sum_{k=1}^n\left(k+\frac{1}{\omega}\right)\left(k+\frac{1}{\omega^2}\right)=340$,then $n=$

The roots of the equation $x^4 - 1 = 0$ are