Cubes which are Sum of Five Cubes

Theorem

The following cube numbers can be expressed as the sum of $5$ positive cube numbers:

$9^3, \ldots$

This article is complete as far as it goes, but it could do with expansion.
In particular: More terms needed. It seems that:
$4$ and all numbers $> 8$ can be so expressed
only $4, 8, 10, 11, 13$ require repeated cubes
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Proof

$\ds 9^3$

$=$

$\ds 729$

$\ds $

$=$

$\ds 1 + 27 + 64 + 125 + 512$

$\ds $

$=$

$\ds 1^3 + 3^3 + 4^3 + 5^3 + 8^3$

This article is complete as far as it goes, but it could do with expansion.
In particular: Add the proof based on $9^3 = 1^3 + 6^3 + 8^3$ and $6^3 = 3^3 + 4^3 + 5^3$ from Cubes which are Sum of Three Cubes
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding this information.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Expand}} from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Sources

1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $729$
1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $729$