Problem: Given a 2 x 2 x 2 cube made of 8 1 x 1 x 1 blocks. If the cube is painted, then each of the blocks has 3 of its faces painted.

If a 3 x 3 x 3 cube is painted, a given block may have 0, 1, 2 , or 3 of its faces painted.

Count how many of each.

Do the same for a 4 x 4 x 4 cube, a 5 x 5 x 5 cube, and a n x n x n cube.

**Now, let us see an example of a 4 x 4 x 4 cube, then
below is three 4 x 4 x 4 cubes with the cubes that will be painted 1, 2,
and 3 sides.**

After looking at the first 4 cubes, the sides painted look as such:

0 blocks with 0 sides painted.

0 blocks with 1 side painted.

0 blocks with 2 sides painted.

8 blocks with 3 sides painted.

1 blocks with 0 sides painted.

6 blocks with 1 side painted.

12 blocks with 2 sides painted.

8 blocks with 3 sides painted.

8 blocks with 0 sides painted.

24 blocks with 1 side painted.

24 blocks with 2 sides painted.

8 blocks with 3 sides painted.

27 blocks with 0 sides painted.

54 blocks with 1 side painted.

36 blocks with 2 sides painted.

8 blocks with 3 sides painted.

**We attempted to look for a pattern after constructing
the first few cubes and looking at the trends in sides painted for each
cube.**

**For 0 sides painted, we found that if n = (the number
blocks on one row or column of the cube), then (n-2)^3 is the number of
cubes with 0 sides painted.**

**For 1 side painted, we found that the pattern was 6(n
- 2)^2 = the number of cubes with 1 side painted.**

**For a cube with 2 sides painted, we found that the pattern
was 12(n - 2) = the number of cubes with 2 sides painted.**

**For the cubes with 3 sides painted, it will always be
8. The eight painted cubes are the 8 corners of any and all cubes.**

**The spreadsheet below takes you through the first nine
cunes with n x n x n sides.**

**Now, we have found these four values for our cubes with
n sides:**

**looking at the four values, one notices that the values
are the products of a binomial expansion.**

**the expansion of the binomial looks as such**