Solving $\frac{\ 3x}{x^2+x+1}+\frac{2x}{x^2-x+1}=3$ with other approaches 
$$\frac{\ 3x}{x^2+x+1}+\frac{2x}{x^2-x+1}=3$$
$$x=?$$

I solved this problem as follow:
$x=0$ is not a root, we  can divide numerator and denominator of each fraction by $x$:
$$\frac{3}{t+1}+\frac{2}{t-1}=3\quad\quad\text{where $t=x+\frac1x$}$$
$$5t-1=3t^2-3\Rightarrow t=2 , \frac{-1}6$$
Only $x+\frac1x=2$ is acceptable and $x=1$.
I'm looking for other elegant methods to solve this equation.

I don't know if you consider this elegant:
\begin{align}
0&=3-\frac{3x}{x^2+x+1}-\frac{2x}{x^2-x+1}\\
&=\left(1-\frac{3x}{x^2+x+1}\right)+2\left(1-\frac{x}{x^2-x+1}\right)\\
&=\frac{(x-1)^2}{x^2+x+1}+2\frac{(x-1)^2}{x^2-x+1}\\
&=(x-1)^2\left(\frac{1}{x^2+x+1}+\frac{2}{x^2-x+1}\right)
\end{align}
Since $x^2+x+1$ and $x^2-x+1$ are always positive, we can cancel the second part to get $(x-1)^2=0$
