# How many cubes are left after removing the surface of a cube?

$n^{3}$ cubes are glued together to form one solid cube which is then hung in the air. As time proceeds, the most outer layer of this solid cube begins to dissolve and eventually those smaller cubes fall to the ground. Exactly how many?

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Hint: How many cubes are left? – Thomas Andrews Apr 3 '13 at 19:37
Well, I thought $(n-2)^{3}$ be left; so the answer should be $n^{3}-(n-2)^{3}$ – user70800 Apr 3 '13 at 19:57
Yes. You can expand $(n-2)^3$ and writ it in a more closed form. – Thomas Andrews Apr 3 '13 at 20:10
I'll make my comment an answer. – Thomas Andrews Apr 3 '13 at 20:11
Please try to make the title a little more specific and informative. – OC2PS May 10 '13 at 2:20

Lucky for us $n^3$ is the volume of a cube of edge n, meaning the cube formed by gluing together $n^3$ cubes has each edge made up of n small cubes. That is a massive hint.
The dissolved cubes would be the surface area of the cube (adjusted for double counting), i.e. $6n^2$ (surface area for 6 surfaces) - $12n$ (removing double counting for 12 edges) + $8$ (adding back the 8 corners which were double removed in double counting adjustment)
Therefore, the number of little cubes remaining $N = n^3 - (6n^2 - 12n +8)$
$= n^3 - 6n^2 + 12n - 8$