If $\mathrm{w},{\mathrm{w}}^2$ are the complex cube roots of unity,
then ${(\mathrm{a} + \mathrm{b})}^3 + {({\mathrm{aw}}^2 + \mathrm{bw})}^3 + {(\mathrm{aw} + {\mathrm{bw}}^2 )}^3 =$

$\begin{array}{rcl}\mathrm{x}& =& \mathrm{a}+\mathrm{b}, \mathrm{y}={\mathrm{aw}}^2+\mathrm{bw}, \mathrm{z}=\mathrm{aw}+{\mathrm{bw}}^2\end{array}$

$\begin{array}{rcl}\mathrm{x}+\mathrm{y}+\mathrm{z}& =& \mathrm{a}+\mathrm{b}+{\mathrm{aw}}^2+\mathrm{bw}+\mathrm{aw}+{\mathrm{bw}}^2\end{array}$

$\begin{array}{rcl}\mathrm{x}+\mathrm{y}+\mathrm{z}& =& \mathrm{a}\left(1+{\mathrm{w}}^2+\mathrm{w}\right)+\mathrm{b}\left(1+\mathrm{w}+{\mathrm{w}}^2\right)\end{array}$

$\begin{array}{rcl} & \Rightarrow & \mathrm{x}+\mathrm{y}+\mathrm{z}=0\\ \therefore {\mathrm{x}}^3+{\mathrm{y}}^3+{\mathrm{z}}^3& =& 3\mathrm{xyz}\end{array}$

$\begin{array}{rcl} & & \\ {\mathrm{x}}^3+{\mathrm{y}}^3+{\mathrm{z}}^3& =& 3(\mathrm{a}+\mathrm{b}) \left({\mathrm{aw}}^2+\mathrm{bw}\right) \left(\mathrm{aw}+{\mathrm{bw}}^2\right)\\ & =& 3\left(\mathrm{a}+\mathrm{b}\right) \left({\mathrm{a}}^2{\mathrm{w}}^3+\mathrm{ab}\left({\mathrm{w}}^4+{\mathrm{w}}^2\right)+{\mathrm{b}}^2{\mathrm{w}}^3\right)\\ & =& 3\left(\mathrm{a}+\mathrm{b}\right) \left({\mathrm{a}}^2-\mathrm{ab}+{\mathrm{b}}^2\right)\\ & =& 3\left({\mathrm{a}}^3+{\mathrm{b}}^3\right)\end{array}$