Question
If $\omega$ is a complex cube root of unity, show that : $\left(2+\omega+\omega^2\right)^3-\left(1-3 \omega+\omega^2\right)^3=65$

Answer

\begin{aligned}
& \text { L.H.S. }=\left(2+\omega+\omega^2\right)^3-\left(1-3 \omega+\omega^2\right)^3 \\
& =\left[2+\left(\omega+\omega^2\right)\right]^3-\left[-3 \omega+\left(1+\omega^2\right)\right]^3 \\
& =(2-1)^3-(-3 \omega-\omega)^3 \\
& =13-(-4 \omega)^3 \\
& =1+64 \omega^3 \\
& =1+64(1) \\
& =65 \\
& =\text { R.H.S. }
\end{aligned}

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