If $\;$ $x = a + b$, $\;$ $y = a \omega + b \omega^2$ $\;$ and $\;$ $z = a \omega^2 + b \omega$, $\;$ show that $\;$ $x^3 + y^3 + z^3 = 3 \left(a^3 + b^3\right)$ $\;$ where $\omega$ is the complex cube root of unity.
$x= a + b$
$\therefore$ $\;$ $x^3 = \left(a + b\right)^3 = a^3 + b^3 + 3 a^2 b + 3 a b^2$ $\;\;\; \cdots \; (1)$
$y = a\omega + b \omega^2$
$\begin{aligned}
\therefore \; y^3 & = \left(a \omega + b \omega^2\right)^3 \\\\
& = a^3 \omega^3 + b^3 \left(\omega^2\right)^3 + 3 a^2 b \omega^4 + 3 a b^2 \omega^5 \\\\
& = a^3 \omega^3 + b^3 \left(\omega^3\right)^2 + 3 a^2 b \omega^3 \times \omega + 3 a b^2 \omega^3 \times \omega^2 \\\\
& = a^3 + b^3 + 3 a^2 b \omega + 3 ab^2 \omega^2 \;\;\; \cdots \; (2) \;\;\; [\text{Note: } \omega^3 = 1]
\end{aligned}$
$z = a \omega^2 + b \omega$
$\begin{aligned}
\therefore \; z^3 & = \left(a \omega^2 + b \omega\right)^3 \\\\
& = a^3 \left(\omega^2\right)^3 + b^3 \omega^3 + 3 a^2 b \omega^5 + 3 a b^2 \omega^4 \\\\
& = a^3 \left(\omega^3\right)^2 + b^3 \omega^3 + 3 a^2 b \omega^3 \times \omega^2 + 3 a b^2 \omega^3 \times \omega \\\\
& = a^3 + b^3 + 3 a^2 b \omega^2 + 3 a b^2 \omega \;\;\; \cdots \; (3)
\end{aligned}$
$\therefore$ $\;$ We have from equations $(1)$, $(2)$ and $(3)$,
$\begin{aligned}
x^3 + y^3 + z^3 & = a^3 + b^3 + 3 a^2 b + 3 ab^2 \\
& \hspace{1cm} a^3 + b^3 + 3 a^2 b \omega + 3 ab^2 \omega^2 \\
& \hspace{2cm} a^3 + b^3 + 3 a^2 b \omega^2 + 3 ab^2 \omega \\\\
& = 3 a^3 + 3 b^3 + 3 a^2 b \left(1 + \omega + \omega^2\right) + 3 ab^2 \left(1 + \omega^2 + \omega\right) \\\\
& = 3 \left(a^3 + b^3\right) \;\;\; [\because \; 1 + \omega + \omega^2 = 0]
\end{aligned}$
Hence proved.