6.351 Solve the equations $x^{3}-6 x^{2}-39 x-10=0$ and $x^{3}+x^{2}-20 x-50=0$, given that one of the roots of the first equation is twice one of the roots of the second equation.
Solution. Let $x_{0}$ be the root of the second equation, $2 x_{0}$ be the root of the first equation, $f(x)=x^{3}-6 x^{2}-39 x-10, g(x)=x^{3}+x^{2}-20 x-50$. Then $x_{0}$ is the root of the equation $\tilde{f}(x)=0$, where $\tilde{f}(x)=f(2 x)=8 x^{3}-24 x^{2}-78 x-10$. Represent $\tilde{f}(x)$ in the form (3). Then $\eta(x)=-32 x^{2}+82 x+390 \Rightarrow \eta(x)=0$ when $\left[\begin{array}{l}x_{0}=5, \\ x_{0}=-\frac{39}{16} .\end{array}\right.$

By verification, we find that $x_{0}=5$, $f(x)=(x-10)\left(x^{2}+4 x+1\right), g(x)=(x-5)\left(x^{2}+6 x+10\right)$.

Answer: The roots of the first equation are: $x_{1}=10, x_{2,3}=-2 \pm \sqrt{3}$; the only root of the second equation is: $x=5$.