6.354 Solve the equation $8 x^{3}+4 x^{2}-34 x+15=0$, given that two of its roots $x_{1}$ and $x_{2}$ satisfy the relation $2 x_{1}-4 x_{2}=1$.
## Solution.

Let $f(x)=8 x^{3}+4 x^{2}-34 x+15$. Then $f\left(x_{1}\right)=f\left(x_{2}\right)=0 \Rightarrow$

$\Rightarrow f\left(x_{1}\right)-f\left(x_{2}\right)=0 \Leftrightarrow 4\left(x_{1}^{3}-x_{2}^{3}\right)+2\left(x_{1}^{2}-x_{2}^{2}\right)-34\left(x_{1}-x_{2}\right)=0 \Leftrightarrow$

$\Leftrightarrow\left(x_{1}-x_{2}\right)\left(8\left(x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}\right)+4\left(x_{1}+x_{2}\right)-34\right)=0 \Rightarrow$

$\Rightarrow\left\{\begin{array}{l}8\left(x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}\right)+4\left(x_{1}+x_{2}\right)-34=0, \\ 2 x_{1}-4 x_{2}=1\end{array} \Rightarrow 28 x_{2}^{2}+16 x_{2}-15=0 \Leftrightarrow\right.$

$\Leftrightarrow\left[\begin{array}{l}x_{2}=\frac{1}{2}, \\ x_{2}=-\frac{15}{4}\end{array} \Rightarrow x_{2}=\frac{1}{2}, f\left(-\frac{15}{4}\right) \neq 0 \Rightarrow x_{1}=\frac{3}{2} \cdot x_{3}=-\frac{1}{2}-x_{2}-x_{1}=-\frac{5}{2}\right.$.

Answer: $x_{1}=\frac{3}{2}, x_{2}=\frac{1}{2}, x_{3}=-\frac{5}{2}$.