Sum of Sequence of Cubes/Also presented as

Theorem
The Sum of Sequence of Cubes can also be presented as:
 * $\ds \sum_{i \mathop = 0}^n i^3 = \paren {\sum_{i \mathop = 1}^n i}^2 = \frac {n^2 \paren {n + 1}^2} 4$

This is seen to be equivalent to the given form by the fact that the first term evaluates as $\dfrac {0^2 \paren {0 + 1}^2 } 4$ which is zero.