(10) The solution set of the inequality $\frac{8}{(x+1)^{3}}+\frac{10}{x+1}-x^{3}-5 x>0$ is
$10(-\infty,-2) \cup(-1,1)$ Hint: Transform the original inequality 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 know that $f(t)$ is monotonically increasing on $\mathbf{R}$, and when $x \neq-1$, the original inequality can be transformed into $f\left(\frac{2}{x+1}\right)>f(x)$, thus $\frac{2}{x+1}>x$, which is equivalent to $\left\{\begin{array}{l}x+1>0, \\ 2>x^{2}+x\end{array}\right.$ or $\left\{\begin{array}{l}x+1<0, \\ 2<x^{2}+x,\end{array}\right.$ Solving these, we get $-1<x<1$ or $x<-2$, hence the solution set of the original inequality is $(-\infty,-2) \cup(-1,1)$.