<s>[INST] 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.
 [/INST] 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.
</s>