10. The solution set of the inequality
$$
\frac{8}{(x+1)^{3}}+\frac{10}{x+1}-x^{3}-5 x>0
$$

is $\qquad$ .
10. $(-\infty,-2) \cup(-1,1)$.

The original inequality is transformed into
$$
\left(\frac{2}{x+1}\right)^{3}+5 \cdot \frac{2}{x+1}>x^{3}+5 x \text {. }
$$

Let $f(t)=t^{3}+5 t$. It is easy to see that $f(t)$ is monotonically increasing on $\mathbf{R}$, and when $x \neq-1$, the original inequality becomes
$$
f\left(\frac{2}{x+1}\right)>f(x) \text {. }
$$

Thus, $\frac{2}{x+1}>x \Leftrightarrow\left\{\begin{array}{l}x+1>0, \\ 2>x^{2}+x\end{array}\right.$ or $\left\{\begin{array}{l}x+1<0, \\ 2<x^{2}+x .\end{array}\right.$
Solving this, we get $-1<x<1$ or $x<-2$.
Therefore, the solution set of the original inequality is $(-\infty,-2) \cup(-1,1)$.