6.357 Solve the equation $3 x^{3}+2 \sqrt{3} x^{2}-21 x+6 \sqrt{3}=0$, given that the product of two of its roots is 1.
Solution. Let $x_{1}, x_{2}, x_{3}$ be the roots of the original equation, $x_{1} x_{2}=1$. By Vieta's theorem, we have:

$\left\{\begin{array}{l}x_{1}+x_{2}+x_{3}=-\frac{2 \sqrt{3}}{3}, \\ x_{1} x_{2} x_{3}=-2 \sqrt{3}\end{array} \Rightarrow\left\{\begin{array}{l}x_{1}+x_{2}=\frac{4 \sqrt{3}}{3}, \\ x_{1} x_{2}=1, \\ x_{3}=-2 \sqrt{3} .\end{array} \Rightarrow x_{1}=\sqrt{3}, x_{2}=\frac{1}{\sqrt{3}}, x_{3}=-2 \sqrt{3}\right.\right.$.

Answer: $x_{1}=\sqrt{3}, x_{2}=\frac{\sqrt{3}}{3}, x_{3}=-2 \sqrt{3}$.