<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] If you are comfortable with generalized harmonic number, you could consider first the partial sum
$$S_n=\sum_{r=1}^{n} \dfrac{r^2 - 1}{r^4 + r^2 + 1}$$ and write first
$$\dfrac{r^2 - 1}{r^4 + r^2 + 1}=\frac{(r-1)(r+1)}{(r-a)(r-b)(r-c)(r-d)}$$
Using partial fraction decomposition, this is
$$\frac{a^2-1}{(a-b) (a-c) (a-d) (r-a)}+\frac{b^2-1}{(b-a) (b-c)
   (b-d) (r-b)}+$$ $$\frac{c^2-1}{(c-a) (c-b) (c-d)
   (r-c)}+\frac{d^2-1}{(d-a) (d-b) (d-c) (r-d)}$$ and use
$$\sum_{r=1}^n \frac 1{r-k}=H_{n-k}-H_{-k}$$
After simplification
$$S_n=\frac{(n-1) n}{2 \left(n^2+n+1\right)}$$
</s>