6.358 Solve the equations $x^{3}-7 x^{2}+12 x-10=0$ and $x^{3}-10 x^{2}-2 x+20=0$, given that one of the roots of the first equation is half of one of the roots of the second equation.
Solution. Let $x_{0}$ be the root of the first equation, $2 x_{0}$ be the root of the second equation, $f(x)=x^{3}-7 x^{2}+12 x-10, g(x)=x^{3}-10 x^{2}-2 x+20$. Then $x_{0}$ is the root of the equation $\tilde{f}(x)=0$, where $\tilde{f}(x)=f(2 x)=8 x^{3}-28 x^{2}+24 x-10$.

Represent $\tilde{f}(x)$ in the form (3). Then $r_{1}(x)=4 x^{2}-25 x+25 ; r_{1}(x)=0 \Leftrightarrow$ $\Leftrightarrow\left[\begin{array}{l}x_{0}=5, \\ x_{0}=\frac{5}{4}\end{array} \Rightarrow x_{0}=5 \Rightarrow f(x)=(x-5)\left(x^{2}-2 x+2\right)\right.$ $g(x)=(x-10)(x-\sqrt{2})(x+\sqrt{2})$

Answer: root of the first equation: $x=5$;

roots of the second equation: $x_{1}=10, x_{2}=-\sqrt{2}, x_{3}=\sqrt{2}$.