Solve for $m$: $(m-4)^3 = \left(\frac 18\right)^{-1}$.
So I can rewrite the equation as $(m-4)^3 = 8$. To get rid of the cube on the left-hand side, I can take the cube root of both sides of the equation. The cube root of $(m-4)^3$ is just $m-4$, and the cube root of $8$ is $2$, since $2^3 = 8$. So I get $m-4 = 2$. Adding $4$ to both sides, I get $m = 6$.

