# Cube as Difference between Two Squares

 It has been suggested that this page or section be merged into Cube Number as Difference between Squares of Triangular Numbers. (Discuss)

## Theorem

A cube number can be expressed as the difference between two squares.

## Proof

 $\ds n^3$ $=$ $\ds \sum_{k \mathop = 1}^n k^3 - \sum_{k \mathop = 1}^{n - 1} k^3$ $\ds$ $=$ $\ds \paren {\frac {n^2 \paren {n + 1}^2} 4} - \paren {\frac {\paren {n - 1}^2 n^2} 4}$ Sum of Sequence of Cubes $\ds$ $=$ $\ds \frac {n^2 \paren {\paren {n + 1}^2 - \paren {n - 1}^2} } 4$ $\ds$ $=$ $\ds \paren {\frac {n \paren {n + 1} } 2}^2 - \paren {\frac {n \paren {n - 1} } 2}^2$

$\blacksquare$