Talk:Cube which can be Represented as Sum of 3, 4, 5, 6, 7 or 8 Cubes

Even if the cubes are required to be distinct, $72^3$ would suffice (there may be smaller):

and also:

These are generated by $3^3 + 4^3 + 5^3 = 6^3$.

If the cubes used in all expressions are required to be distinct, $216^3$ would suffice (again, there may be smaller):

I am not sure why $351120$ is so special. --RandomUndergrad (talk) 13:58, 31 August 2020 (UTC)


 * Likewise.
 * Either Wells completely misunderstood and misconstrued something he read somewhere and turned something profound into something trivial, or it's a misprint of some kind. I think it's time for a Historical Note. --prime mover (talk) 14:07, 31 August 2020 (UTC)