Hello, here is nice solution. You can also factorize this:
$\sin^3 18^{\circ}+\sin^2 18^{\circ}=\sin^2 18^{\circ}(\sin 18^{\circ}+1)=\sin^2 18^{\circ}(\sin 18^{\circ}+\sin 90^{\circ})=$,
now we can use formula $\sin x+\sin y=2\sin(\frac{x+y}{2})\cos(\frac{x-y}{2})$ to get $\sin 18^{\circ}+\sin 90^{\circ}=2\sin 54^{\circ} \cos 36^{\circ}=2\cos ^2 36^{\circ}$, so:
$\sin^3 18^{\circ}+\sin^2 18^{\circ}=2\sin^2 18^{\circ} \cos^2 36^{\circ}=\frac{2\sin^2 18^{\circ} \cos^2 36^{\circ} \cos^2 18^{\circ}}{\cos^218^{\circ}}=\frac{\sin^2 36^{\circ} \cos^2 36^{\circ}}{2 \cos^2 18^{\circ}}$, and now we can use formula $\sin^2 x \cdot \cos^2x=\frac{\sin^2 2x}{4}$, so $\sin^2 36^{\circ} \cos^2 36^{\circ}=\frac{\sin^2 72 ^{\circ}}{4}$, and finally:
$\sin^3 18^{\circ}+\sin^2 18^{\circ}=\frac{\sin^2 72 ^{\circ}}{8\cos^2 18^{\circ}}=\frac{1}{8}$