### Prove that, ${\sin ^3}30^\circ - {\sin ^3}18^\circ = {\sin ^2}18^\circ$

We have, ${\sin ^3}18^\circ + {\sin ^2}18^\circ = {\sin ^2}18^\circ (\sin 18^\circ + 1)$

$= {\left( {\frac{{\sqrt 5 - 1}}{4}} \right)^2}\left( {\frac{{\sqrt 5 - 1}}{4} + 1} \right)$

$= \frac{{6 - 2\sqrt 5 }}{{16}} \times \frac{{\sqrt 5 + 3}}{4}$

$= \frac{{3 - \sqrt 5 }}{8} \times \frac{{3 + \sqrt 5 }}{4}$

$= \frac{{9 - 5}}{{8 \times 4}} = \frac{1}{8} = {\sin ^3}30^\circ$

$\Rightarrow {\sin ^2}18^\circ = {\sin ^3}30^\circ - {\sin ^3}18^\circ$