Solve for $x$: $(x-4)^3=\left(\frac18\right)^{-1}$
I can start by simplifying the right-hand side, since it has a negative exponent. Using the property that $a^{-b}=\frac1{a^b}$, I can rewrite the right-hand side as $\left(\frac18\right)^{-1}=8^1=8$. Now the equation is $(x-4)^3=8$. To undo the cube, I can take the cube root of both sides. The cube root of 8 is 2, since $2^3=8$. The cube root of $(x-4)^3$ is $x-4$, since the cube and the cube root are inverse operations. So I have $x-4=2$. To solve for $x$, I just need to add 4 to both sides. That gives me $x=2+4=6$. 