Sum of Sequence of Cubes/Proof by Recursion

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
Let $\displaystyle A \left({n}\right) = 1 + 2 + \cdots + n = \sum_{i \mathop = 1}^n i = \frac{n \left({n + 1}\right)} 2$.

Let $\displaystyle B \left({n}\right) = 1^2 + 2^2 + \cdots + n^2 = \sum_{i \mathop = 1}^n i^2 = \frac{n \left({n + 1}\right) \left({2 n + 1}\right)} 6$.

Let $\displaystyle S \left({n}\right) = 1^3 + 2^3 + \cdots + n^3 = \sum_{i \mathop = 1}^n i^3$.