Sum of Sequence of Cubes

Theorem

 * $\displaystyle \sum_{i \mathop = 1}^n i^3 = \paren {\sum_{i \mathop = 1}^n i}^2 = \frac {n^2 \paren {n + 1}^2} 4$

Historical Note
This result was documented by in his work Āryabhaṭīya of $499$ CE.