Olympiad problem with cubes

[TomCowley]

<font color="white">It's obvious that a cube can be dissected into n^3 equal cubes. It's also obvious that n^3 of these equal cubes, aligned correctly, can be combined into one cube. Inductively, if 55-61 exist, then breaking any cube into 8 subcubes will generate all further numbers.

So the operations +7, +26, +63, +124, and +215 always exist, and the operations -7, -26, -63, etc. conditionally exist.

55: Split into a 3x3x3 cube (27), split 4 into 2x2x2 (+28)

56: Split into a 6x6x6 cube (216), condense 4 sets of 27 cubes into 3x3x3 cubes (-104 = 112) (leaving a 6x6x3 set of cubes on the bottom and 4 big 3x3x3 cubes on top), then condense 8 sets of 8 cubes (-56 = 56)

57: 64 - 7

58: 216 - 5*26 -4*7

59: 64 - 26 + 7*3

60: 27 + 26 + 7

61: 27 + 4*26 (blow up a 2x2 square on one side) - 10*7 (9 from the subsubcubes and 1 from the subcubes)

There's probably a proof based on a +3*26 - 11*7 being an increase of one cube, or someting close to that, but I'm not sure how to show 55's relevance in any clean way (my solution for 61 doesn't have room for another -7, and 216 - 3*26 - 11*7 also doesn't have room for another -7).</font>