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

## 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.

$\blacksquare$

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.