Expressing the surface area of a cube as a function of its volume is like plugging in the surface area into volume? Analyzing $\text{S.A.}=6s^2$, and $V=s^3$ I would assume that the Surface area is plugged into the volume. So then $$s=\sqrt{\frac{\text{SA}}{6}}$$ and the volume would $$V=\left(\frac{\text{SA}}{6}\right)^{3/2}$$
$\endgroup$ 32 Answers
$\begingroup$What you have is volume as a function of surface area. You want the other way around, meaning you'll need to invert what you have: $$ \textsf{SurfaceArea} = 6(\textsf{Volume})^{2/3} $$
$\endgroup$ $\begingroup$One sixth of surface area = $a^2 =(a^3)^{2/3}=V^{2/3}$
$\endgroup$
