3. If $a>b>0$, then the minimum value of $a^{3}+\frac{1}{b(a-b)}$ is $\qquad$ .


$$
\begin{array}{l}
\text { 3. } \frac{5}{3} \sqrt[5]{144} . \\
a^{3}+\frac{1}{b(a-b)} \geqslant a^{3}+\frac{4}{a^{2}}=2 \times \frac{a^{3}}{2}+3 \times \frac{4}{3 a^{2}} \\
\geqslant 5 \sqrt[5]{\left(\frac{1}{2}\right)^{2} \times\left(\frac{4}{3}\right)^{3}}=\frac{5}{3} \sqrt[5]{144} .
\end{array}
$$

The equality holds if and only if $\left\{\begin{array}{l}b=a-b, \\ \frac{a}{2}=\frac{4}{3 a^{2}},\end{array}\right.$ which is $\left\{\begin{array}{l}a=\sqrt[3]{\frac{8}{3}}, \\ b=\frac{1}{2} \sqrt[3]{\frac{8}{3}}\end{array}\right.$.
