Question 552409
Your picture did not show. It must have looked something like this, but with 9 squares on each face of the cube
{{{drawing(300,240,-5,5,-1,7,
rectangle(-4,0,0,4),
line(0,0,3.464,2),line(0,2,3.464,4),line(0,4,3.464,6),
line(-2,4,1.464,6),line(-4,4,-0.536,6),
line(-4,2,0,2),line(-2.268,5,1.732,5),line(-0.536,6,3.464,6),
line(-2,4,-2,0),line(1.732,5,1.732,1),line(3.464,2,3.464,6)
)}}} And my picture should look like one of the 2 by 2 by 2 cubes found in the large cube.
A given face of a cube has 4 neighbor faces, sharing an edge. Only the 6th face has no common edge; it is opposite and parallel to the first (given) face. So, in any cube with exactly 3 red faces, at least two of the red faces must share an edge.
In 2 by 2 by 2 cubes cut from the painted 3 by 3 by 3 cube, the 3 red faces must share are vertex, because we cannot have two red faces parallel to each other. So the 2 by 2 by 2 cubes must include a corner/vertex of the original painted 3 by 3 by 3 cube. Since there are only 8 such corners, there will be only 8 different 2 by 2 by 2 cubes with exactly 3 red faces.