6 Let $\alpha \in\left(0, \frac{\pi}{2}\right)$, then the minimum value of $\frac{\sin ^{3} \alpha}{\cos \alpha}+\frac{\cos ^{3} \alpha}{\sin \alpha}$ is ( ).
(A) $\frac{27}{64}$
(B) $\frac{3 \sqrt{2}}{5}$
(C) 1
(D) $\frac{5 \sqrt{3}}{6}$
6
$$
\begin{aligned}
\frac{\sin ^{3} \alpha}{\cos \alpha}+\frac{\cos ^{3} \alpha}{\sin \alpha} & =\frac{\sin ^{4} \alpha+\cos ^{4} \alpha}{\sin \alpha \cos \alpha} \\
& =\frac{1-2(\sin \alpha \cos \alpha)^{2}}{\sin \alpha \cos \alpha} .
\end{aligned}
$$

Let $t=\sin \alpha \cos \alpha=\frac{1}{2} \sin 2 \alpha \in\left(0, \frac{1}{2}\right]$, then
$$
f(t)=\frac{1-2 t^{2}}{t}=\frac{1}{t}-2 t
$$

is monotonically decreasing in $t \in\left(0, \frac{1}{2}\right]$. Therefore,
$$
f(t)_{\min }=f\left(\frac{1}{2}\right)=1,
$$

which means $\frac{\sin ^{3} \alpha}{\cos \alpha}+\frac{\cos ^{3} \alpha}{\sin \alpha}$ has a minimum value of 1. Hence, the answer is C.