If $\alpha, \beta$ are the roots of the equation $x^2+x+1=0$,then $(\alpha+\beta)^2+(\alpha^2+\beta^2)^2+(\alpha^3+\beta^3)^2+\ldots+(\alpha^{12}+\beta^{12})^2=$

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 $\omega$ is a complex cube root of unity,then $(x+1)(x+\omega)(x-\omega-1)$ is equal to

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 $\frac{1}{A_3}+\frac{1}{A_6}+\frac{1}{A_9}+\frac{1}{A_{12}}+\cdots \infty =$

Let $\alpha, \beta$ be the roots of the equation $x^2-\sqrt{6}x+3=0$ such that $\operatorname{Im}(\alpha)>\operatorname{Im}(\beta)$. Let $a, b$ be integers not divisible by $3$ and $n$ be a natural number such that $\frac{\alpha^{99}}{\beta}+\alpha^{98}=3^n(a+ib)$,where $i=\sqrt{-1}$. Then $n+a+b$ is equal to:

If $\left(\frac{\sqrt{3}+i}{\sqrt{3}-i}\right)^4+\left(\frac{\sqrt{3}-i}{\sqrt{3}+i}\right)^4=r \operatorname{cis} \theta$,then one of the values of $\sqrt{r \operatorname{cis} \theta}$ is