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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 $

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