An exploded-view diagram of a cube broken into interior, faces, edges, and corners.


A while ago I arbitrarily decided that I needed a favourite three-digit number (don’t ask) and ended up choosing 216. It’s a nice cube number — 6×6×6 — and can also be expressed as a sum of three smaller cubes:

6^3 = 5^3 + 4^3 + 3^3

The Wikipedia article for the number has a diagram showing one way to reassemble a 6×6×6 cube into three smaller cubes, but I’ve been playing around looking for other, more aesthetically pleasing methods. Here’s one I found.

First, we break the 6×6×6 cube into its 4×4×4 interior, six 4×4×1 faces, twelve 4×1×1 edges, and eight 1×1×1 corners1: i.e.,

6^3 = 4^3 + 6\cdot 4^2 + 12\cdot 4^1 + 8 \cdot 4^0
An exploded-view diagram of a cube separated into an interior, faces, edges, and corners.
Decomposing the cube according to its polytope boundaries.

The 4×4×1 faces can be combined with seven of the edges and one of the corners to build a 5×5×5 cube. The remaining five edges can be split into ten 2×1×1 chunks and arranged with the remaining seven corners to form a 3×3×3 cube.

An exploded-view diagram of a 3×3×3 cube and a 5×5×5 cube using the pieces of the 6×6×6 cube from before.
Rearranging the pieces into smaller cubes.
The unexploded view of the 3×3×3, 4×4×4, and 5×5×5 cubes made from the pieces of the 6×6×6 cube.
The final three cubes.

There are many more ways to construct three cubes from the pieces of a 6×6×6 cube. What’s your favourite?

  1. As an aside, this decomposition can be done with any size cube and even in any number of dimensions. An n-dimensional hypercube of size x≥2 breaks into k-dimensional polytope boundary chunks as xn = ∑​ 2nk C(n, k) (x−2)k , where C(n, k) is the binomial coefficient. ↩︎