jmc

The value $x=f^{-1}(-31/96)$ is the solution to $f(x)=-31/96$. This means $$\frac{x^5-1}{3}=\frac{-31}{96}.$$Multiplying by 3 gives $$x^5-1=\frac{-31}{32}.$$If we add 1 we get $$x^5=\frac{-31}{32}+\frac{32}{32}=\frac{1}{32},$$and the only value that solves this equation is $$x=\boxed{\frac{1}{2}}.$$