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