If $1, \omega, \omega^2$ are the cube roots of unity, then…
If $1, \omega, \omega^2$ are the cube roots of unity, then $(2-\omega)^2\left(2-\omega^2\right)^2\left(2-\omega^{10}\right)^2\left(2-\omega^{11}\right)^2=$
$-7^4$
$7^4$
$7^8$
$-7^8$
Solution
$1, \omega, \omega^2$ are the cube roots of units.
$\Rightarrow 1+\omega+\omega^2=0$
Given,
$(2-\omega)^2\left(2-\omega^2\right)^2\left(2-\omega^{10}\right)^2\left(2-\omega^{11}\right)^2$
$\begin{aligned} & =\left\{(2-\omega)\left(2-\omega^2\right)\right\}^2\left\{(2-\omega)\left(2-\omega^2\right)\right\}^2 \\ & =\left\{(2-\omega)\left(2-\omega^2\right)\right\}^4 \\ & =\left\{4-2\left(\omega+\omega^2\right)+\omega^3\right\}^4 \\ & =(4-2(-1)+1)^4=7^4\end{aligned}$