jmc

Factor $x^2+2x-15$ as $(x+5)(x-3)$ to find that $r$ is either $3$ or $-5$. Before evaluating $\frac{r^3-1}{r^5+r^4-r^3-r^2}$ at each of these values, we first simplify it. The numerator factors as a difference of cubes, and the denominator has a common factor of $r^2$. $$\frac{r^3-1}{r^5+r^4-r^3-r^2}=\frac{(r-1)(r^2+r+1)}{r^2(r^3+r^2-r-1)}.$$The expression in parentheses in the denominator factors as $$r^3+r^2-r-1=r^2(r+1)-(r+1)=(r^2-1)(r+1),$$so the original expression is $$\frac{(r-1)(r^2+r+1)}{r^2(r^2-1)(r+1)}=\frac{(r-1)(r^2+r+1)}{r^2(r-1)(r+1)(r+1)}=\frac{(r^2+r+1)}{r^2(r+1{)}^2}.$$Evaluating this expression at $r=3$ and $r=-5$ yields $13/144$ and $21/400$ respectively. Therefore, the maximum value of the expression is $\boxed{\frac{13}{144}}$.