jmc

We note that $j(x)$ is defined unless one or more of the denominators $x+8,x^2+8,x^3+8$ is equal to $0$.

We have $x+8=0$ if $x=-8$, and $x^3+8$ if $x=\sqrt[3]{-8}=-2$. There is no real $x$ for which $x^2+8=0$. Therefore, the domain of $j(x)$ consists of all real $x$ except $-8$ and $-2$. As a union of intervals, this is $\boxed{(-\infty ,-8)\cup (-8,-2)\cup (-2,\infty )}$.