China Mathematical Competition (Shaanxi)

Notice that $|x|=3$ is a root of the equation $|x{|}^3-2x^2-4|x|+3=0$. Then the original inequality can be rewritten as $(|x|-3)(|x{|}^2+|x|-1)<0$, that is $$(|x|-3)\left(|x|-\frac{-1+\sqrt{5}}{2}\right)\left(|x|-\frac{-1-\sqrt{5}}{2}\right)<0.$$ Since $|x|-\frac{-1-\sqrt{5}}{2}>0$, then $\frac{-1+\sqrt{5}}{2}<|x|<3$. So the solution set is $(-3,-\frac{\sqrt{5}-1}{2})\cup (\frac{\sqrt{5}-1}{2},3)$.