9. The solution set of the inequality $|x|^{3}-2 x^{2}-4|x|+3<0$ is $\qquad$ .
9. $\left(-3,-\frac{\sqrt{5}-1}{2}\right) \cup\left(\frac{\sqrt{5}-1}{2}, 3\right)$.

The original inequality can be transformed into
$$
|x|^{3}-2|x|^{2}-4|x|+30.
$$
Thus, (1) $\Rightarrow(|x|-3)\left(|x|-\frac{\sqrt{5}-1}{2}\right)<0$
$\Rightarrow \frac{\sqrt{5}-1}{2}<|x|<3$
$\Rightarrow-3<x<-\frac{\sqrt{5}-1}{2}$ or $\frac{\sqrt{5}-1}{2}<x<3$.
Therefore, the solution set of the inequality is
$$
x \in\left(-3,-\frac{\sqrt{5}-1}{2}\right) \cup\left(\frac{\sqrt{5}-1}{2}, 3\right) .
$$