Solve for $m$: $(m-4)^3 = \left(\frac 18\right)^{-1}$.
We have $\left(\frac{1}{8}\right)^{-1}=8=2^3$, so we can write the given equation as $$(m-4)^3=2^3.$$ Therefore, $m-4 = 2$, so $m=\boxed{6}$.
6
Level 2
Algebra
Problem: Solve for $m$: $(m-4)^3 = \left(\frac 18\right)^{-1}$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve for $m$, first simplify the right side of the equation. The expression $\left(\frac{1}{8}\right)^{-1}$ can be rewritten using the property of exponents that states $a^{-b} = \frac{1}{a^b}$. Thus, $\left(\frac{1}{8}\right)^{-1} = 8$. 

Now, the equation becomes $(m-4)^3 = 8$. To isolate $m-4$, take the cube root of both sides. The cube root of $8$ is $2$, so we have $m-4 = 2$. 

Next, solve for $m$ by adding $4$ to both sides: $m = 2 + 4$. This simplifies to $m = 6$. 

Therefore, the final answer is $6$.<|END_OF_LATENT|> The final answer is $6$.