8. $x^{2}+x+1$, $x^{2}-1$, $2x+1$ are the three sides of a triangle, the largest interior angle of this triangle is $\qquad$ .
8. $120^{\circ} \cdot \cos \theta=\frac{\left(x^{2}-1\right)^{2}+(2 x+1)^{2}-\left(x^{2}+x+1\right)^{2}}{2\left(x^{2}-1\right)(2 x+1)}=-\frac{1}{2}$, so $\theta=120^{\circ}$.