Sum of five cubes

It is true that all integers can be expressed as the sum of cubes of five integers, not necessarily distinct.

Note the identity $6k=(k+1)^3+(k-1)^3-k^3-k^3$ We use the above identity for $k=\large{\frac{n^3-n}{6}=\frac{n(n-1)(n+1)}{6}}$ which is an integer for all $n$ as the product of 3 consecutive numbers is divisible by $3!=6$ .

This gives us $n^3-n=\left(\frac{n^3-n}{6}+1\right)^3+\left(\frac{n^3-n}{6}-1\right)^3-\left(\frac{n^3-n}{6}\right)^3-\left(\frac{n^3-n}{6}\right)^3$

Therefore, $\boxed{n=(-n)^3+\left(\frac{n-n^3}{6}-1\right)^3+\left(\frac{n-n^3}{6}+1\right)^3+\left(\frac{n^3-n}{6}\right)^3+\left(\frac{n^3-n}{6}\right)^3}$ .