9.041. $x^{6}-9 x^{3}+8>0$.

Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.

9.041. $x^{6}-9 x^{3}+8>0$.
Solution.

Let $x^{3}=y$. Then we get $y^{2}-9 y+8>0$. Solving this inequality, we find $\left[\begin{array}{l}y>8, \\ y<1.\end{array}\right.$

Since $y = x^3$, we have $\left[\begin{array}{l}x^{3}>8, \\ x^{3}<1.\end{array}\right.$

This can be rewritten as $\left[\begin{array}{l}x>2, \\ x<1.\end{array}\right.$

Since $x^{3}-1=(x-1)\left(x^{2}+x+1\right)$ and $x^{2}+x+1>0$ for $x \in \mathbb{R}$, the last system of inequalities is equivalent to the system $\left[\begin{array}{l}x-2>0, \\ x-1<0.\end{array}\right.$

Answer: $\quad x \in(-\infty ; 1) \cup(2 ; \infty)$.