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 $