jmc

Seeing the expression $\sqrt[3]{x}$ twice, we make the substitution $y=\sqrt[3]{x},$ so that our inequality becomes $$y+\frac{2}{y+3}\le 0.$$Combining the terms on the left-hand side under a common denominator, we get $$\frac{y^2+3y+2}{y+3}\le 0,$$which factors as $$\frac{(y+1)(y+2)}{y+3}\le 0.$$Letting $f(y)=(y+1)(y+2)/(y+3),$ we make a sign table based on this inequality: \begin{array}{c|ccc|c} &$y+1$ &$y+2$ &$y+3$ &$f(y)$ \\ \hline$y<-3$ &$-%%DISP_0%%amp;$-%%DISP_0%%amp;$-%%DISP_0%%amp;$-$\\ [.1cm]$-3<y<-2$ &$-%%DISP_0%%amp;$-%%DISP_0%%amp;$+%%DISP_0%%amp;$+$\\ [.1cm]$-2<y<-1$ &$-%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$-$\\ [.1cm]$y>-1$ &$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+$\\ [.1cm]\end{array}Therefore, the inequality holds if $y<-3$ or $-2<y<-1.$ Since the inequality is nonstrict, we must also include the values of $y$ that make $f(y)=0,$ which are $y=-1$ and $y=-2.$ Therefore, the solutions to this inequality are $$y\in (-\infty ,-3)\cup [-2,-1].$$Since $y=\sqrt[3]{x},$ we have either $\sqrt[3]{x}<-3$ or $-2\le \sqrt[3]{x}\le -1.$ Since $\sqrt[3]{x}$ is an increasing function of $x,$ we can cube all sides of these inequalities, to get $x<-27$ and $-8\le x\le -1,$ respectively. Therefore, $$x\in \boxed{(-\infty ,-27)\cup [-8,-1]}.$$