8. The equation $x^{3}+t x+s=0\left(t \in \mathbf{R}, s \in \mathbf{C}, \arg s=\frac{\pi}{6}\right)$ about $x$ has 3 complex roots, which correspond to the 3 vertices of an equilateral triangle with side length $\sqrt{3}$ in the complex plane. Find the values of $s$ and $t$.
8. Let the complex number corresponding to the center of the equilateral triangle be $z_{0}$, and the complex numbers corresponding to the 3 roots be $z_{1}, z_{2}, z_{3}$, and
$$
\omega=\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3} \text {, }
$$

then $z_{2}-z_{0}=\left(z_{1}-z_{0}\right) \omega, z_{3}-z_{0}=\left(z_{1}-z_{0}\right) \omega^{2}$.
By Vieta's formulas, we have
$$
\begin{array}{l}
z_{1}+z_{2}+z_{3}=\left(1+\omega+\omega^{2}\right) z_{1}+\left(2-\omega-\omega^{2}\right) z_{0}=3 z_{0}=0, \\
t=z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}=\left(\omega+\omega^{2}+\omega^{3}\right) z_{1}^{2}=0, \\
s=-z_{1} z_{2} z_{3}=-z_{1}^{3} .
\end{array}
$$

Also, $\left|z_{1}\right|=\left|z_{1}-z_{0}\right|=1$, then $|s|=1$.
Thus, $s=\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}=\frac{\sqrt{3}}{2}+\frac{1}{2} i$.