Integer Solutions of the Equation $u^3 = r^2-s^2$ The question says the following:

Find all primitive Pythagorean Triangles $x^2+y^2 = z^2$ such that $x$ is a perfect cube.

The general solution for each variable are the following:
$$x=r^2-s^2$$
$$y=2rs$$
$$z=r^2+s^2$$
such that $\gcd(r,s) = 1$and $r+s \equiv 1 \pmod {2}$
In order to make $x$ a perfect cube, I shall have the equation $x=u^3=r^2-s^2$. However, I am stuck to find a general formula for such cubes. 
I know that a subset of the solutions might be the difference between two consecutive squares. This difference is always an odd integer. I can collect some examples such $14^2-13^2 = 27$ but I cannot give a formula for such type either.
Any ideas?

$1^2=1^3$
$3^2=(1+2)^2=1^3+2^3$
$6^2=(1+2+3)^2=1^3+2^3+3^3$
...
The difference between two consecutive squares on the left will give you a cube:

$1^3=1^2-0^2$
$2^3=3^2-1^2$
$3^3=6^2-3^2$
...
Which means the solutions are pairs of this form: $(\frac{n(n-1)}{2}, \frac{n(n+1)}{2})$ 
$1^3=1^2-0^2$
$2^3=3^2-1^2$
$3^3=6^2-3^2$
...
