Notes: `(-1/2 + sqrt(3)/2 i )^3` Use the Binomial Theorem to expand the complex number, then simplify the result.

The expansion of $\displaystyle{{\left({x}+{y}\right)}}^{{n}}$ is,

$\displaystyle_{n}{C}_{{0}}{{x}}^{{n}}+_{{n}}{C}_{{1}}{{x}}^{{{n}-{1}}}{{y}}^{{1}}+_{{n}}{C}_{{2}}{{x}}^{{{n}-{2}}}{{y}}^{{2}}+\ldots\ldots.+_{{n}}{C}_{{n}}{{y}}^{{n}}$

$\displaystyle={{x}}^{{n}}+{n}{{x}}^{{{n}-{1}}}{y}+\frac{{{n}!}}{{{2}!{\left({n}-{2}\right)}!}}{{x}}^{{{n}-{2}}}{{y}}^{{2}}+\ldots\ldots+{{y}}^{{n}}$

$\displaystyle\therefore{{\left(-\frac{{1}}{{2}}+\frac{\sqrt{{{3}}}}{{2}}{i}\right)}}^{{3}}{\)}={{\left(-\frac{{1}}{{2}}\right)}}^{{3}}+{3}{{\left(-\frac{{1}}{{2}}\right)}}^{{{3}-{1}}}\cdot{\left(\frac{\sqrt{{{3}}}}{{2}}{i}\right)}+_{{3}}{C}_{{2}}\cdot{{\left(-\frac{{1}}{{2}}\right)}}^{{{3}-{2}}}\cdot{{\left(\frac{\sqrt{{{3}}}}{{2}}{i}\right)}}^{{2}}+{{\left(\frac{\sqrt{{{3}}}}{{2}}{i}\right)}}^{{3}}$

$\displaystyle=-\frac{{1}}{{8}}+{3}{{\left(-\frac{{1}}{{2}}\right)}}^{{2}}\cdot{\left(\frac{\sqrt{{{3}}}}{{2}}{i}\right)}+\frac{{{3}!}}{{{2}!{\left({3}-{2}\right)}!}}\cdot{{\left(-\frac{{1}}{{2}}\right)}}^{{1}}\cdot{{\left(\frac{\sqrt{{{3}}}}{{2}}{i}\right)}}^{{2}}+{{\left(\frac{\sqrt{{{3}}}}{{2}}{i}\right)}}^{{3}}$

$\displaystyle=-\frac{{1}}{{8}}+{\left(\frac{{{3}\sqrt{{{3}}}}}{{8}}\right)}{i}-\frac{{3}}{{2}}\cdot{\left(\frac{{{3}{{i}}^{{2}}}}{{4}}\right)}+\frac{{{3}\sqrt{{{3}}}{{i}}^{{3}}}}{{8}}$