Represent the cube with the 3x3x3 integer lattice, with edges between adjacent points. Now it’s clear that to separate the points we have to cut all the edges. The 9 corner points have degree 3, the 12 edge points have degree 4, the 6 face points have degree 5 and the center has degree 6. So there are (36 + 48 + 30 +6)/2 = 54 edges.

If we only allow plane cuts, then by inspection it’s easy to see that we can cut at most 9 edges at once. So 54/9 = 6 cuts are required.

Note though that if we can cut weird shapes then you can do this in fewer cuts (presumably this is not allowed by the “buzzsaw”). ]]>

[spoiler]He suggested a visualization where you paint the outside of the cube before cutting. Of the final 27 cubes, some will have three painted faces, some two, etc. Painted faces are “free” as they already existed but each of the non-painted faces required a cut to accomplish. As you point out, the inner cube has six unpainted faces.[/spoiler]

]]>[spoiler]Without thinking through all the possible dissections, let’s consider the “inner” cube. No matter how you slice things, that one cube has no existing cut faces and will require you to make individual cuts for each of its 6 new faces.

Hence, it can’t be done with less than 6 cuts.[/spoiler]

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