4. Given an isosceles triangle with a vertex angle of $20^{\circ}$ and a base length of $a$, the length of the legs is $b$. Then the value of $\frac{a^{3}+b^{3}}{a b^{2}}$ is $\qquad$
4.3.

Given $a=2 b \sin 10^{\circ}$.
Thus $a^{3}+b^{3}=8 b^{3} \sin ^{3} 10^{\circ}+b^{3}$
$$
\begin{array}{l}
=8 b^{3} \cdot \frac{1}{4}\left(3 \sin 10^{\circ}-\sin 30^{\circ}\right)+b^{3}=6 b^{3} \sin 10^{\circ} \\
\Rightarrow \frac{a^{3}+b^{3}}{a b^{2}}=\frac{6 b^{3} \sin 10^{\circ}}{2 b \sin 10^{\circ} \cdot b^{2}}=3 .
\end{array}
$$