Evaluate $\sum_{r=1}^{\infty} \dfrac{r^2 - 1}{r^4 + r^2 + 1}$ I was only able to observe that:
$\dfrac{r^2 - 1}{r^4 + r^2 + 1} = \dfrac{r^2 - 1}{(r^2 + r + 1)(r^2 - r + 1)}$
This hints at telescoping, but I would need an $r$ term in the numerator.

The original question was

Evaluate $\sum_{r=1}^{\infty} \dfrac{r^3 + (r^2 + 1)^2}{(r^4 + r^2 + 1)(r^2 + r)}$

I was able to simplify it to the following:
$\dfrac{r^3 + (r^2 + 1)^2}{(r^4 + r^2 + 1)(r^2 + r)} = \dfrac{(r^4 + r^3 - r^2 - r)}{(r^4 + r^2 + 1)(r^2 + r)} + \dfrac{3r^2+r+1}{\{(r+1)(r^2 + r + 1)\}\{r(r^2 - r + 1)\}} = \dfrac{r^2 - 1}{r^4 + r^2 + 1} + \left[\dfrac{1}{r(r^2 - r + 1)} - \dfrac{1}{(r+1)(r^2 + r + 1)}\right]$
The second term can be evaluated using telescoping, and the first term is what this post is asking for.
Any other ways of solving the original question are also welcome.

As this Art of Problem Solving thread suggests, write this using a  form of partial fractions:
$$(Ar + B)(r^2 - r  + 1) + (Cr + D)(r^2 + r + 1) = r^2 - 1$$
Now comparing the coefficients of $r^3$, $r^2$, $r$, and the constant term, we end up with $A + C = 0, B - A + D + C = 1, -B + A+D+C=0$, and $B + D = -1$.
Adding the middle two equations gives $D + C = 1/2$, and so $B - A = 1/2$. Now $B + (C) = 1/2$ and so we obtain $2(B + C + D) = 0$, which leads to $B = -1/2$, $C = 1$, $D = -1/2$, and $A = -1$.
Hence the summand can be written as:
$$\frac{-r - 1/2}{r^2 + r + 1} + \frac{r-1/2}{r^2 - r + 1}$$
which is telescoping.
