Earlier quoted context omitted.
The greedy algorithm works here: 2017 = 12^3 + 6^3 + 4^3 + 2^3 + 1^3
Technical note: 5 cubes is not enough for every number, so this property of 2017 is not trivial. From > Every positive integer can be written as the sum of nine (or fewer) positive cubes. This upper limit of nine cubes cannot be reduced because, for example, 23 cannot be written as the sum of fewer than nine positive cubes: > 23 = 2^3 + 2^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3. I couldn't find if it's common tha…
5, 12, 19, 26, 31, 33, 38, 40, 45, 52, 57, 59, 64, 68, 71, 75, 78, 82, 83, 89, 90, 94, 96, 97, 101, 108, 109, 115, 116, 120, 127, 129, 131, 134, 135, 136, 138, 143, 145, 146, 150, 152, 153, 155, 157, 162, 164, 169, 171, 172, 176, 181, 183, 188, 190, 192, 194
It seems this is fairly common (1757 is the 1000th such number), but of course that says nothing.
Reading http://mathworld.wolfram.com/CubicNumber.html, it is true that every sufficiently large integer is a sum of no more than 7 positive cubes.
It also states ”the only integers requiring nine positive cubes are 23 and 239. Wieferich proved that only 15 integers require eight cubes: 15, 22, 50, 114, 167, 175, 186, 212, 231, 238, 303, 364, 420, 428, and 454 (OEIS A018889).”
Even stronger (same page): ”Deshouillers et al. (2000) conjectured that 7373170279850 is the largest integer that cannot be expressed as the sum of four nonnegative cubes” (nice title for a paper: ”7 373 170 279 850.”. See http://www.ams.org/journals/mcom/2000-69-229/S0025-5718-99-0...)
If that is true, it is indeed common that 5 cubes is enough (since 4 almost always would be sufficient)