We approached the problem individually at first, then we began to compare notes. We first agreed that N could not be 1 because that cube would have 6 painted sides
Some started making a table for cubes of increasing dimension: 2*2*2, 3*3*3, 4*4*4, etc.They counted cubes in each of the four categories for each size cube, filled in the table and looked for a pattern.
Others took an algebraic approach:
if n was some number not equal to 1, the cube would have 8 corners with 3 ptd fcs;
each of the 12 edges would minus the corners would have (n-2) cubes with ptd fcs;
each of the 6 exposed surfaces (minus corners and edges) would have (n-2)^2 cubes with ptd fcs;
all the inner unpainted cubes would for a cube with side (n-2) for a total of (n-2)^3 small cubes.
The general form for the total num,ber of cubes works out to be: n^3 - 4n^2 + 18n - 28
We plugged values from the table into the algebraic formula and found that both methods produced the same results for several different cubes. We assumed the others would correspond,also.