Let this common value be $r$. Then, for all $1\leq i\leq 2009$, we have that $$x_i^2 - x_ix_{i+1} + x_{i+1}^2 = r,$$with indices taken mod $2009$, so $$\frac{x_i^3+x_{i+1}^3}{x_i+x_{i+1}} = r,$$(since $x_i + x_{i+1} > 0$ we can do this), so $$(x_i^3-rx_i) + (x_{i+1}^3-rx_{i+1}) = 0.$$Let $y_i = x_i^3-rx_i$, then $y_i + y_{i+1} = 0$ for all $1\leq y\leq 2009$, so $y_i = y_{i+2}$, so since $\gcd(2, 2009) = 1$ and indices are taken mod $2009$, it follows that all $y_i$ are equal. Since $y_i + y_{i+1}=0$, it then follows that all $y_i$ are $0$. Thus, $x_i^3 - rx_i = 0$ for all $i$. Since $x_i > 0$, $x_i \neq 0$, so $x_i^2 - r = 0$ for all $i$. Clearly $r > 0$ as $r = x_1^2-x_1x_2+x_2^2 = \frac{3}{4}(x_1-x_2)^2+\frac{1}{4}(x_1+x_2)^2 > 0$. So, it follows that $x_i = \sqrt{r}$ for all $i$, i.e. all $x_i$'s are equal.