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.

Write \begin{align}&\frac{r^2-1}{r^4+r^2+1} = \frac{Ar+B}{r^2-r+1}+\frac{Cr+D}{r^2+r+1}\\ \iff &r^2-1 = (Ar+B)(r^2+r+1)+(Cr+D)(r^2-r+1)\tag{1}\end{align}
You can now expand the RHS as a polynomial in $r$ and by direct comparison of coefficients on the LHS and the RHS get a $4\times4$ linear system.
Alternatively, let $\alpha$ be a root of $x^2-x+1$. In particular, we have $\alpha^2 = \alpha - 1$. Evaluating $(1)$ at $\alpha$ gives us
\begin{align}\alpha^2-1 &= (A\alpha + B)(\alpha^2+\alpha+1)\\
(\alpha-1)-1 &= (A\alpha + B)((\alpha-1)+\alpha+1)\\
\alpha - 2 &= 2\alpha(A\alpha+B)\\
\alpha - 2 &= 2A\alpha^2+2B\alpha\\
\alpha - 2 &= 2A(\alpha - 1)+2B\alpha\\
\alpha - 2 &= 2(A+B)\alpha-2A \implies 2(A+B)=1,\ -2A = -2
\end{align}
Note that it's important to get rid of any higher powers of $\alpha$ besides $\alpha$ and $1$ before you compare coefficients. Similarly, if you take $\beta$ a root of $x^2+x+1$, you can calculate $C$ and $D$ analogously.
Now that you can calculate partial fractions decomposition, we have
\begin{align}
\sum_{r = 1}^n \frac{r^2-1}{r^4+r^2+1} &=\sum_{r = 1}^n\left( \frac{2r-1}{2(r^2-r+1)}-\frac{2r+1}{2(r^2+r+1)} \right) \\
&= \sum_{r = 1}^n \frac{2r-1}{2(r^2-r+1)} - \sum_{r = 1}^n \frac{2r+1}{2(r^2+r+1)}\\
&= \sum_{r = 1}^n \frac{2r-1}{2(r^2-r+1)} - \sum_{r = 1}^n \frac{2(r+1)-1}{2((r+1)^2-(r+1)+1)} \\
&= \sum_{r = 1}^n \frac{2r-1}{2(r^2-r+1)} - \sum_{r = 2}^{n+1} \frac{2r-1}{2(r^2-r+1)} \\
&= \frac 12 - \frac{2n+1}{2(n^2+n+1)}
\end{align}
Taking a limit as $n\to \infty$ gives us $$\sum_{r = 1}^\infty \frac{r^2-1}{r^4+r^2+1} =  \frac 12.$$
