It is taken that n is integral, because otherwise the cube would be exceedingly difficult, if not impossible, to divide in the aforesaid way.
a) No matter how long the side, only the corner unit cubes will have three sides of paint on them, because this is the most number of corners revealed for any unit cube in a cube with side larger than 1. There are always eight corner unit cubes (in cubes with sides larger than 1), and so the formula is f(n) = 8, where f(n) --> 3 painted sides, unless n = 1, where f(n) = 0. (There is only one cube for a unit cube itself, with 6 painted sides).
b) The edges of the cube, not including the corners, will have two of their sides painted, those cubes having only two sides revealed. There are twelve edges for each cube. However, each edge of n includes the two corner cubes, which must be subtracted because of their painting of three sides. Thus, g(n) = 12(n-2), where n is larger than 1 (for reasons explained already in a)).
c) The center unit cubes, of each face, will only have one surface painted, because this is the only one revealed. These center cubes do not include those of the edge, or the corners. Thus, there are n-2 by n-2 of these, for each face, and six faces of the large cube. h(n) = 6 * (n-2)^2, where n is larger than 1 (for reasons explained already in a)).
d) The remaining cubes are not painted, as the most exposed unit cube in a cubical structure would only be the corner, revealed on three sides. These three, two, and one sides have already been covered, and so the remainder are unpainted unit cubes. There are n^3 total unit cubes present, and so the number of these unpainted unit cubes is j(n) = n^3 - 6 * (n-2)^2 - 12(n-2) - 8, where n is necessarily larger than 1 (for the above explained reasons, in a)).
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