Reindex a rational series by two methods

Given $\sum_{r=2}^n\dfrac{3r^2+1}{(r^2-1)^3}=\dfrac12\left[\dfrac98-\dfrac1{n^3}-\dfrac1{(n+1)^3}\right]$, find $\sum_{r=3}^n\dfrac{3(r-1)^2+1}{[(r-1)^2-1]^3}$ in terms of $n$. Hence find its infinite sum.