Sum of Sequence of Cubes

Theorem

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

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