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

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

\(\ds 72^3\) \(=\) \(\ds 36^3 + 48^3 + 60^3\)
\(\ds \) \(=\) \(\ds 3^3 + 18^3 + 29^3 + 70^3\)
\(\ds \) \(=\) \(\ds 18^3 + 24^3 + 30^3 + 48^3 + 60^3\)
\(\ds \) \(=\) \(\ds 3^3 + 9^3 + 12^3 + 15^3 + 29^3 + 70^3\)
\(\ds \) \(=\) \(\ds 9^3 + 12^3 + 15^3 + 24^3 + 30^3 + 48^3 + 60^3\)
\(\ds \) \(=\) \(\ds 3^3 + 6^3 + 8^3 + 9^3 + 10^3 + 15^3 + 29^3 + 70^3\)

and also:

\(\ds 72^3\) \(=\) \(\ds 1^3 + 6^3 + 8^3 + 12^3 + 15^3 + 24^3 + 30^3 + 48^3 + 60^3\)
\(\ds \) \(=\) \(\ds 12^3 + 15^3 + 16^3 + 20^3 + 21^3 + 28^3 + 32^3 + 35^3 + 40^3 + 57^3\)
\(\ds \) \(=\) \(\ds 1^3 + 3^3 + 4^3 + 5^3 + 8^3 + 12^3 + 15^3 + 24^3 + 30^3 + 48^3 + 60^3\)
\(\ds \) \(=\) \(\ds 6^3 + 8^3 + 10^3 + 15^3 + 16^3 + 20^3 + 21^3 + 28^3 + 32^3 + 35^3 + 40^3 + 57^3\)

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):

\(\ds 216^3\) \(=\) \(\ds 48^3 + 76^3 + 212^3\)
\(\ds \) \(=\) \(\ds 96^3 + 100^3 + 159^3 + 161^3\)
\(\ds \) \(=\) \(\ds 15^3 + 20^3 + 25^3 + 164^3 + 178^3\)
\(\ds \) \(=\) \(\ds 18^3 + 24^3 + 30^3 + 77^3 + 166^3 + 171^3\)
\(\ds \) \(=\) \(\ds 13^3 + 51^3 + 68^3 + 78^3 + 85^3 + 104^3 + 195^3\)
\(\ds \) \(=\) \(\ds 21^3 + 39^3 + 52^3 + 65^3 + 110^3 + 115^3 + 126^3 + 168^3\)

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)
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