A cube is painted (all 6 faces) and cut into one-inch cubes. If the original cube has n inch edges, how many (in terms of n) one-inch cubes have:a. exactly 3 faces painted?b. exactly 2 faces painted?c. exactly 1 face painted?d. exactly 0 faces painted?e. What is the total number of one-inch cubes?

Answer:a. 8b. 12(n -2)c. 6(n -2)^2d. (n -2)^3Step-by-step explanation:A cube has 8 vertices (corners), 12 edges, and 6 faces.a. Three faces will be painted on each of the 8 corner cubes.__b. Two faces will be painted on the edge cubes that are not corner cubes. For an n-inch edge, n-2 inches of it are not part of the corner cubes. Two faces will be painted on 12(n -2) edge cubes.__c. One face will be painted on each face cube that is not part of an edge or a corner. For a face of dimensions n inches square, the cubes of interest comprise a square that is n-2 inches on a side. There are 6 such faces of the larger cube. One face will be painted on 6(n -2)^2 face cubes.__d. The cubes internal to the larger cube that are not part of any face, edge, or corner make up a cube that is (n -2)^3 smaller cubes. This is the number with no paint.__e. Of course the total number of cubes with or without paint is n^3. This is the sum of the cubes in each category:   (((n -2) +6)(n -2) +12)(n -2) +8 = ((n +4)(n -2) +12)(n -2) +8   = (n^2 +2n +4)(n -2) +8   = n^3 +2n^2 -2n^2 +4n -4n -8 +8   = n^3