Sum of Sequence of Cubes/Proof by Nicomachus

Proof
By Nicomachus's Theorem, we have:


 * $\forall n \in \N_{>0}: n^3 = \left({n^2 - n + 1}\right) + \paren {n^2 - n + 3} + \ldots + \paren {n^2 + n - 1}$

Also by Nicomachus's Theorem, we have that the first term for $\paren {n + 1}^3$ is $2$ greater than the last term for $n^3$.

So if we add them all up together, we get:

Hence the result.