Sum of Sequence of Cubes/Proof by Nicomachus

Theorem

 * $\displaystyle \sum_{i \mathop = 1}^n i^3 = \left({\sum_{i \mathop = 1}^n i}\right)^2 = \frac{n^2 \left({n + 1}\right)^2} 4$

Proof
By Nicomachus's Theorem, we have:


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

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

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

Hence the result.