If $\frac{1}{x} + x = 2\cos \theta,$ then ${x^n} + \frac{1}{{{x^n}}}$ is equal to

If $x$ is a cube root of unity other than $1$,then $\left(x+\frac{1}{x}\right)^2+\left(x^2+\frac{1}{x^2}\right)^2+\ldots+\left(x^{12}+\frac{1}{x^{12}}\right)^2=$

If the number of real roots of $x^9-x^5+x^4-1=0$ is $n$,the number of complex roots having argument on the imaginary axis is $m$,and the number of complex roots having argument in the $2^{nd}$ quadrant is $k$,then $m \cdot n \cdot k = $

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} = $

If $\alpha, \beta$ are the roots of the equation $x^2-2x+4=0$,then $\alpha^n+\beta^n = \ldots \cos \left(\frac{n\pi}{3}\right)$ for any $n \in N$.

If $\omega = \frac{-1 + \sqrt{3}i}{2}$,then $(3 + \omega + 3\omega^2)^4 = $