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.

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
$$S_n=\frac{(n-1) n}{2 \left(n^2+n+1\right)}$$
