An algebra problem by Ankit Nigam

If $\omega^3 = 1$ , $\Rightarrow \omega^2+\omega+1 = 0$ this is because multiplying the latter equation with $\omega$ , we have $\omega^3+\omega^2+\omega = 0$ . Add both side with $1$ , $\Rightarrow \omega^3+\omega^2+\omega +1 = 1$ , $\Rightarrow \omega^3+0 = 1$ .

Therefore,

$\begin{aligned} (1+\omega-\omega^2)^7 & = (-\omega^2-\omega^2)^7= (-2\omega^2)^7 = -128\omega^{14} = -128\omega^{12}\omega^2 \\ & = -128(\omega^3)^4\omega^2 = -128(1)^4\omega^2 = \boxed{-128\omega^2} \end{aligned}$