Compute $\sin^3 18^\circ + \sin^2 18^\circ.$
We can write
\begin{align*}
\sin^3 18^\circ + \sin^2 18^\circ &= \sin^2 18^\circ (\sin 18^\circ + 1) \\
&= \sin^2 18^\circ (\sin 18^\circ + \sin 90^\circ).
\end{align*}By sum-to-product,
\begin{align*}
\sin^2 18^\circ (\sin 18^\circ + \sin 90^\circ) &= \sin^2 18^\circ \cdot 2 \sin 54^\circ \cos 36^\circ \\
&= 2 \sin^2 18^\circ \cos^2 36^\circ \\
&= \frac{2 \sin^2 18^\circ \cos^2 18^\circ \cos^2 36^\circ}{\cos^2 18^\circ} \\
&= \frac{4 \sin^2 18^\circ \cos^2 18^\circ \cos^2 36^\circ}{2 \cos^2 18^\circ}.
\end{align*}Then by double angle formula,
\begin{align*}
\frac{4 \sin^2 18^\circ \cos^2 18^\circ \cos^2 36^\circ}{2 \cos^2 18^\circ} &= \frac{\sin^2 36^\circ \cos^2 36^\circ}{2 \cos^2 18^\circ} \\
&= \frac{4 \sin^2 36^\circ \cos^2 36^\circ}{8 \cos^2 18^\circ} \\
&= \frac{\sin^2 72^\circ}{8 \cos^2 18^\circ} \\
&= \boxed{\frac{1}{8}}.
\end{align*}Alternatively, we can plug in the value $\sin 18^\circ = \frac{\sqrt{5} - 1}{4}.$