Answer
$10.94\approx11$ hours.
Work Step by Step
Step 1. Using the volume-to-side relation: $V=s^3$, we have $dV/dt=3s^2\ ds/dt$
Step 2. With the given formula $\frac{dV}{dt}=-k(6s^2), k\gt0$, we have $3s^2\frac{ds}{dt}=-k(6s^2)$ or $\frac{ds}{dt}=-2k$, which is a constant.
Step 3. Thus, we can model the function of $s(t)$ with a line $s(t)=-2kt+s_0$ where $s_0$ is the starting side length.
Step 4. Using the given conditions $V(0)=V_0=s_0^3, s_0=\sqrt[3] V_0$, and $V(1)=\frac{3}{4}V_0$, we have $s^3(1)=\frac{3}{4}V_0$ and $s(1)=\sqrt[3] {\frac{3}{4}V_0}=\sqrt[3] {\frac{3}{4}}s_0$
Step 5. Combine the above results to get $\sqrt[3] {\frac{3}{4}}s_0=-2k+s_0$ and $2k=s_0(1-\sqrt[3] {\frac{3}{4}})$
Step 6. For the cube to melt completely, let $s(t)=0$; we have $0=-2kt+s_0$, thus $t=\frac{s_0}{2k}=\frac{1}{1-\sqrt[3] {\frac{3}{4}}}\approx10.94\approx11$ hours.
