Prove that if $\omega^3 = 1$, then $\left(\dfrac{-1 + i \sqrt{3}}{2}\right)^5 + \left(\dfrac{-1 - i \sqrt{3}}{2}\right)^5 = -1$
Given: $\;$ $\omega^3 = 1$
$\implies$ $\omega$ is the cube root of unity.
Then, $\omega = \dfrac{-1 + i \sqrt{3}}{2}$ $\;$ and $\;$ $\omega^2 = \dfrac{-1 - i \sqrt{3}}{2}$
Now,
$\begin{aligned}
\left(\dfrac{-1 + i \sqrt{3}}{2}\right)^5 + \left(\dfrac{-1 - i \sqrt{3}}{2}\right)^5 & = \left(\omega\right)^5 + \left(\omega^2\right)^5 \\\\
& = \omega^3 \times \omega^2 + \left(\omega^3\right)^3 \times \omega \\\\
& = \omega^2 + \omega \;\;\; \left[\because \; \omega^3 = 1\right] \\\\
& = -1 \;\;\; \left[\because \; 1 + \omega + \omega^2 = 0\right]
\end{aligned}$
Hence proved.