Let $x = \alpha + \beta$,$y = \alpha \omega + \beta \omega^2$,and $z = \alpha \omega^2 + \beta \omega$,where $\omega$ is an imaginary cube root of unity. The product $xyz$ is equal to:

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 the cube roots of unity are $1, \omega, \omega^2$,then the roots of the equation $(x - 2)^3 + 27 = 0$ are

If $1, \omega, \omega^2$ are the cube roots of unity and $(x+y)(x \omega+y \omega^2)(x \omega^2+y \omega)=f(x, y)$,then $f(2, 3)=$

If $1, \omega, \omega^2, \omega^3, \dots, \omega^{n-1}$ are the $n^{th}$ roots of unity,then $(1 - \omega)(1 - \omega^2) \dots (1 - \omega^{n-1})$ equals

If $\omega$ is a complex cube root of unity,then $225+(3 \omega+8 \omega^2)^2+(3 \omega^2+8 \omega)^2$ is equal to: