jmc

Move every term to the left-hand side and factor to find $$\begin{array}{rl}{x}^3(x^2-x-72)& =0\\ x^3(x-9)(x+8)& =0\\ x^3(x-9)(x+8)& =0.\end{array}$$Setting $x^3=0$, $x-9=0$ and $x+8=0$, we find that there are $\boxed{3}$ solutions: $x=0$, $x-9$ and $x=-8$.