6. Let $a>b>0$. Then the minimum value of $a^{3}+\frac{12}{b(a-b)}$ is $\qquad$ .


6.20
$$
\begin{array}{l}
a^{3}+\frac{12}{b(a-b)} \geqslant a^{3}+\frac{12}{\frac{(b+a-b)^{2}}{4}} \\
=a^{3}+\frac{48}{a^{2}}=\frac{a^{3}}{2}+\frac{a^{3}}{2}+\frac{16}{a^{2}}+\frac{16}{a^{2}}+\frac{16}{a^{2}} \geqslant 20 .
\end{array}
$$