> cubes have 8 corners, but I am having trouble visualizing them coming out of the six faces like your describing Yeah that was some bad extrapolating. Uncle G

> cubes have 8 corners, but I am having trouble visualizing them coming out of the six faces like your describing Yeah that was some bad extrapolating. Uncle Google gave me a helpful picture [1] but I'll try and work through the math the right way: Starting with two cubes: two of the faces of the base cube will have prisms and four of the faces will have a tetrahedron. Adding a third cube you can either have the prism faces on the same side or not. The former gives two faces with 5 volumes (prisms intersecting) and 4 faces with 7 (tetrahedrons intersecting). The latter gives 4 faces with 5 volumes (tetrahedron intersecting prism) and 2 faces with 7 volumes (tetrahedrons intersecting). So go with prisms on the same side. This tells us there's 38 volumes completely outside C1 - 32 truly exterior pieces that are only part of C2 or C3 and 6 interior where C2 and C3 overlap on each face. Repeating this process for C2 and C3 will count all the truly exterior pieces twice 32 * 3 / 2 = 48 exterior pieces. There are 6 volumes for each combination of 2 cubes so 6 * C(3,2) = 18 intersections of two cubes. And there's one dodecahedron in the center where all 3 cubes overlap. 48 + 18 + 1 = 67 which appears to be the correct answer according to quora. [1] https://en.wikipedia.org/wiki/Compound_of_three_cubes https://en.wikipedia.org/wiki/Compound_of_three_cubes