The volume of a cube with edge length ee is given by:

\(\displaystyle{V}={e}^{{3}}\)

Differentiate with respect to t: \(\frac{dV}{dt}=3e^2(\frac{de}{dt})\)

Substitute \(\frac{dV}{dt}=1200\ \frac{cm^3}{min}\) and \(e=20\) cm then solve for \(\frac{de}{dt}:\) \(\displaystyle{1200}={3}{\left({20}\right)}^{{2}}\cdot{\left({}\frac{{de}}{{\left.{d}{t}\right.}}\right)}\)

\(\displaystyle{1200}={1200}{\left({}\frac{{de}}{{\left.{d}{t}\right.}}\right)}\)

\(\displaystyle{1}={}\frac{{de}}{{\left.{d}{t}\right.}}\)

or

\(\displaystyle{}\frac{{de}}{{\left.{d}{t}\right.}}={1}{}\frac{{cm}}{\min}\)