6.343 Let $x_{1}, x_{2}, x_{3}$ be the roots of the equation $x^{3}+p x^{2}+q x+r=0$.

1) Form the equation with roots $x_{1} x_{2}, x_{2} x_{3}, x_{1} x_{3}$;
2) use the result from part 1 to find the roots of the equation $x^{3}-3 \sqrt{2} x^{2}+7 x-3 \sqrt{2}=0$.
Solution. 1) By Vieta's theorem, we have: $\left\{\begin{array}{l}x_{1}+x_{2}+x_{3}=-p, \\ x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}=q, \\ x_{1} \cdot x_{2} \cdot x_{3}=-r .\end{array}\right.$

By the inverse of Vieta's theorem, the desired equation is:

$y^{3}-\left(x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}\right) y^{2}+\left(x_{1}^{2} x_{2} x_{3}+x_{2}^{2} x_{1} x_{3}+x_{3}^{2} x_{1} x_{2}\right) y-\left(x_{1} x_{2} x_{3}\right)^{2}=0 \Leftrightarrow$ $\Leftrightarrow y^{3}-q y^{2}+x_{1} x_{2} x_{3}\left(x_{1}+x_{2}+x_{3}\right) y-r^{2}=0 \Leftrightarrow y^{3}-q y^{2}+r p y-r^{2}=0$.

2) Let $x_{1}, x_{2}, x_{3}$ be the roots of the equation $x^{3}-3 \sqrt{2} x^{2}+7 x-3 \sqrt{2}=0$.

Then $x_{1} x_{2}, x_{2} x_{3}, x_{1} x_{3}$ are the roots of the equation $y^{3}-7 y^{2}+18 y-18=0 \Leftrightarrow$ $\Leftrightarrow(y-3)\left(y^{2}-4 y+6\right)=0 \Rightarrow y_{1}=3, y_{2} y_{3}=6$, where $y_{1}=x_{1} x_{2}, y_{2}=x_{2} x_{3}$, $y_{3}=x_{1} x_{3}$. Hence, $x_{1} x_{2}=3, x_{1} x_{2} x_{3}^{2}=6 \Rightarrow x_{3}^{2}=2 \Rightarrow x_{3}=\sqrt{2}$.

The other roots are complex.

Answer: $x=\sqrt{2}$.