We know from the isoperimetric inequality that locally the surface must be a sphere (where we can include the plane as the limiting case of a sphere with infinite radius). Also, the surface must be orthogonal to the cube where they meet; if they're not, you can deform the surface locally to reduce its area. A sphere orthogonal to a cube face must have its centre on that face. You can easily show that it can't contain half the volume if it intersects only one or two faces. Thus, it must either intersect at least three adjacent faces, in which case its centre has to be at the vertex where they meet, or it has to intersect at least two opposite faces, in which case it has to be a plane.
I think we're better off showing it a different way. First, we show that a square has a smaller perimeter than any other rectangle of the same area. Consider a rectangle of sides $s+t$ and $s-t$, where $s > t > 0$. Its area is $s^2-t^2$. A square of the same area would have a side of $\sqrt{s^2-t^2} < \sqrt{s^2} < s$, so its perimeter would be smaller than that of the rectangle.
Now, we move to the three-dimensional case. Suppose we had a rectangular solid of length $l$, width $w$, and depth $d$, with $l > w$. Consider the cross-section of area $lw$. We claim that a rectangular solid of greater volume can be generated by using a square cross-section of side $s$. By the above lemma, we know that we can create a cross-section of equal area $s^2 = lw$, but with perimeter $4s < 2(l+w)$. Thus a solid of dimensions $s$ by $s$ by $d$ would have the same volume as the original solid, but less surface area. By the same token, however, we could make the solid deeper than $d$, to restore the original surface area, but increase the volume.
By repeating the same argument with width and depth (instead of length and width), we establish that a rectangular solid of whatever dimensions always has a smaller volume than a cube of identical surface area.
The above argument is far from rigorous, but it can be made so, and maybe it's convincing enough for your purposes?
Best Answer
I think what you mean is that you have a cube of $\textbf{volume}$ $ \ 8\text{cm}^3$=2cm x 2cm x 2cm.
A cube has 6 faces of equal area.
Therefore, the total surface area of the cube is equal to 6 multiplied by the area of one of the faces.
The area of one of the faces is 4$\text{cm}^2$ = 2cm x 2cm.
Therefore the total area of the cube is 24$\text{cm}^2$ = 6 x 4$\text{cm}^2$.
One thing you must remember is your units of measurement, writing 8cm2 (2x2x2) really makes no sense, since you are equating an area with a single number without units.
The result of your internet search has returned the method for working out the volume of the cube (or in fact a cuboid).
Volume = Length x Width x Height $ \ \ $ (cm$^3$ = cm x cm x cm)