9. If $0<a<\sqrt{3} \sin \theta, \theta \in\left[\frac{\pi}{4}, \frac{5 \pi}{6}\right]$, then the minimum value of $f(a, \theta)=\sin ^{3} \theta+\frac{4}{3 a \sin ^{2} \theta-a^{3}}$ is $\qquad$
9.3.

Since $3 a \sin ^{2} \theta-a^{3}=\frac{1}{\sqrt{2}} \cdot \sqrt{2 a}\left(3 \sin ^{2} \theta-a^{2}\right)=\frac{1}{\sqrt{2}} \sqrt{2 a^{2}\left(3 \sin ^{2} \theta-a^{2}\right)^{2}}$ $\leqslant \frac{1}{\sqrt{2}}\left[\frac{2\left(3 \sin ^{2} \theta-a^{2}\right)+2 a^{2}}{3}\right]^{\frac{3}{2}}=2 \sin ^{3} \theta$. Therefore, $f(a, \theta) \geqslant \sin ^{3} \theta+\frac{2}{\sin ^{3} \theta}$, given $\theta \in\left[\frac{\pi}{4}, \frac{5 \pi}{6}\right]$, we have $\frac{1}{8} \leqslant \sin ^{2} \theta \leqslant 1$. Since the function $y=x+\frac{2}{x}$ is monotonically decreasing in $\lfloor 0, \sqrt{2}\rfloor$, it follows that $f(a, \theta) \geqslant \sin ^{3} \theta+\frac{2}{\sin ^{3} \theta} \geqslant 3$. The equality holds when $\theta=\frac{\pi}{2}, a=1$.