8. The equation $\mathrm{z}^{8}+\mathrm{z}^{3}+1=0$ has a complex root, and the argument of this root on the complex plane is between $90^{\circ}$ and $180^{\circ}$. Find the degree of the argument.
Let $\omega=z^{3}$, the given equation transforms into $\omega^{2}+\omega+1=0$. Its solutions are $-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ and $\frac{-1-\sqrt{3}i}{2}$, with arguments of $120^{\circ}$ and $240^{\circ}$, respectively. Since $z=\sqrt[3]{\omega}$, we can find the six arguments of $z$ to be $\frac{120^{\circ}}{3}, \frac{120^{\circ}+360^{\circ}}{3}, \frac{120^{\circ}+720^{\circ}}{3}$, $\frac{240^{\circ}}{3}, \frac{240^{\circ}+360^{\circ}}{3}, \frac{240^{\circ}+720^{\circ}}{3}$. Clearly, only the second one, which is $160^{\circ}$, lies between $90^{\circ}$ and $180^{\circ}$, and is the required value of $\theta$ in degrees.