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.

$$r^4+r^2+1=(r^2+1)^2-r^2=(r^2+r+1)(r^2-r+1)$$
Observe that if $f(n)=n^2+n+1, f(r-1)=(r-1)^2-(r-1)+1=r^2-r+1$
If $g(n)=an+b, g(n-1)=a(n-1)+b=an+b-a$
$$\text{Let  }\dfrac{r^2-1}{r^4+r^2+1}=\dfrac{g(r)}{f(r)}-\dfrac{g(r-1)}{f(r-1)}$$ $$\text{so that the partial sum becomes }\sum_{r=1}^n\dfrac{r^2-1}{r^4+r^2+1}=\dfrac{g(n)}{f(n)}-\dfrac{g(0)}{f(0)}$$
$$\text{Now, }\lim_{n\to\infty}\dfrac{g(n)}{f(n)}=\lim_{n\to\infty}\dfrac{an+b}{n^2+n+1}=\lim_{n\to\infty}\dfrac{\dfrac an+\dfrac b{n^2}}{1+\dfrac1n+\dfrac1{n^2}}=0\text{ for finite a,b}$$
$$\text{ and } \dfrac{g(0)}{f(0)}=?\text{  so we don't need the actual value of }a$$
$$\begin{align}\implies r^2-1=(ar+b)(r^2-r+1)-(ar+b-a)(r^2+r+1)=-ar^2+(a-2b)r+a\end{align}$$
$$\implies -a=1\iff a=-1, a-2b=0\iff b=\dfrac a2=-\dfrac12$$
