cube of an integer

Theorem.  Any cube of integer is a difference of two squares, which in the caseof a positive cube are the squares of two successive triangular numbers.

For proving the assertion, one needs only to check the identity

$$a^3\equiv {\left(\frac{(a+1)a}{2}\right)}^2-{\left(\frac{(a-1)a}{2}\right)}^2.$$

For example we have  ${(-2)}^3=1^2-3^2$  and  $4^3=64={10}^2-6^2$.

Summing the first $n$ positive cubes, the identity allows http://planetmath.org/encyclopedia/TelescopingSum.htmltelescoping between consecutive brackets,

$1^3+2^3+3^3+4^3+\mathrm{\dots }+n^3$$\mathrm{\hspace{0.33em}}=[1^2-0^2]+[3^2-1^2]+[6^2-3^2]+[{10}^2-6^2]+\mathrm{\dots }+\left[{\left({\displaystyle \frac{(n+1)n}{2}}\right)}^2-{\left({\displaystyle \frac{(n-1)n}{2}}\right)}^2\right]$

$\mathrm{\hspace{0.33em}}={\left({\displaystyle \frac{n^2+n}{2}}\right)}^2,$

saving only the square ${\left(\frac{(n+1)n}{2}\right)}^2$.  Thus we have this expression presenting the sum of the first $n$ positive cubes (cf. the Nicomachus theorem).