Find the ordered triple of positive integers $(a,b,c)$ such that \begin{align*}a &= \frac{1908}{b(c-b)}-\frac{1908}{b(c+b)} \\ b &= \frac{2600}{a(c-a)}-\frac{2600}{a(c+a)} \\ c &= \frac{686}{a(b-a)}-\frac{686}{a(b+a)} .\end{align*}