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

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Theorem

$351 \, 120^3$ can be represented as the sum of $3$, $4$, $5$, $6$, $7$ or $8$ cubes.


Proof

\(\displaystyle 351120^3\) \(=\) \(\displaystyle 175560^3 + 234080^3 + 292600^3\)
\(\displaystyle \) \(=\) \(\displaystyle 2 \times 87780^3 + 204820^3 + 321860^3\)
\(\displaystyle \) \(=\) \(\displaystyle 2 \times 87780^3 + 175560^3 + 2 \times 263340^3\)
\(\displaystyle \) \(=\) \(\displaystyle 3 \times 117040^3 + 3 \times 234080^3\)
\(\displaystyle \) \(=\) \(\displaystyle 2 \times 58520^3 + 117040^3 + 3 \times 175560^3 + 292600^3\)
\(\displaystyle \) \(=\) \(\displaystyle 8 \times 175560^3\)

These representations are not necessarily unique.

$\blacksquare$


Additional Results

We also have:

\(\displaystyle 351120^3\) \(=\) \(\displaystyle 58520^3 + 2 \times 117040^3 + 5 \times 175560^3 + 234080^3\)
\(\displaystyle \) \(=\) \(\displaystyle 5 \times 87780^3 + 4 \times 175560^3 + 263340^3\)

So $351120^3$ can be expressed as a sum of $9$ or $10$ cubes.


These equations all stem from:

\(\displaystyle 12^3\) \(=\) \(\displaystyle 6^3 + 8^3 + 10^3\)
\(\displaystyle \) \(=\) \(\displaystyle 2 \times 3^3 + 7^3 + 11^3\)
\(\displaystyle \) \(=\) \(\displaystyle 2 \times 3^3 + 6^3 + 2 \times 9^3\)
\(\displaystyle \) \(=\) \(\displaystyle 3 \times 4^3 + 3 \times 8^3\)
\(\displaystyle \) \(=\) \(\displaystyle 2 \times 2^3 + 4^3 + 3 \times 6^3 + 10^3\)
\(\displaystyle \) \(=\) \(\displaystyle 8 \times 6^3\)
\(\displaystyle \) \(=\) \(\displaystyle 2^3 + 2 \times 4^3 + 5 \times 6^3 + 8^3\)
\(\displaystyle \) \(=\) \(\displaystyle 5 \times 3^3 + 4 \times 6^3 + 9^3\)

showing that $351 \, 120$ is not the smallest number with this property.


Sources