6.356 Solve the equation $x^{4}-6 x^{3}+7 x^{2}+6 x-2=0$, given that it has at least one pair of roots $x_{1}$ and $x_{2}$, the difference of which is 1.
Solution. Let $f(x)=x^{4}-6 x^{3}+7 x^{2}+6 x-2 ; x_{1}-x_{2}=1$, then $f\left(x_{1}\right)=f\left(x_{2}\right)=0 \Rightarrow f\left(x_{1}\right)-f\left(x_{2}\right)=0 \Leftrightarrow$ $\Leftrightarrow\left(x_{1}^{4}-x_{2}^{4}\right)-6\left(x_{1}^{3}-x_{2}^{3}\right)+7\left(x_{1}^{2}-x_{2}^{2}\right)+6\left(x_{1}-x_{2}\right)=0 \Leftrightarrow$ $\Leftrightarrow\left(x_{1}+x_{2}\right)\left(x_{1}^{2}+x_{2}^{2}\right)-6\left(x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}\right)+7\left(x_{1}+x_{2}\right)+6=0 \Leftrightarrow$ $\Leftrightarrow\left(x_{1}+x_{2}\right)\left(\left(x_{1}-x_{2}\right)^{2}+2 x_{1} x_{2}\right)-6\left(\left(x_{1}-x_{2}\right)^{2}+3 x_{1} x_{2}\right)+7\left(x_{1}+x_{2}\right)+6=0 \Leftrightarrow$ $\Leftrightarrow\left(x_{1}+x_{2}\right)\left(1+2 x_{1} x_{2}\right)-6\left(1+3 x_{1} x_{2}\right)+7\left(x_{1}+x_{2}\right)+6=0 \Rightarrow$ $\Rightarrow\left\{\begin{array}{l}\left(x_{1}+x_{2}\right)\left(4+x_{1} x_{2}\right)=9 x_{1} x_{2}, \\ x_{1}=x_{2}+1\end{array} \Rightarrow x_{2}^{3}-3 x_{2}^{2}+2=0 \Leftrightarrow\right.$ $\Leftrightarrow\left(x_{2}-1\right)\left(x_{2}-1-\sqrt{3}\right)\left(x_{2}-1+\sqrt{3}\right)=0$. Since $f(1) \neq 0$, then $\left[\begin{array}{l}x_{2}=1+\sqrt{3}, x_{1}=2+\sqrt{3} \\ x_{2}=1-\sqrt{3}, x_{1}=2-\sqrt{3}\end{array}\right.$

1) $x_{1}=2+\sqrt{3}, x_{2}=1+\sqrt{3} \Rightarrow$ (by Vieta's theorem) $\left\{\begin{array}{l}x_{3}+x_{4}=6-x_{1}-x_{2}, \\ x_{3} x_{4}=-\frac{2}{x_{1} x_{2}}\end{array} \Leftrightarrow\left\{\begin{array}{l}x_{3}+x_{4}=3-\sqrt{3}, \\ x_{3} x_{4}=5-3 \sqrt{3}\end{array} \Leftrightarrow\left[\begin{array}{l}x_{3}=1-\sqrt{3}, x_{4}=2-\sqrt{3}, \\ x_{3}=2-\sqrt{3}, x_{4}=1-\sqrt{3} .\end{array}\right.\right.\right.$
2) $x_{1}=2-\sqrt{3}, x_{2}=1-\sqrt{3}$. This case was obtained by considering point 1.

Answer: The roots of the equation are: $x_{1}=1+\sqrt{3}, x_{2}=2+\sqrt{3}, x_{3}=1-\sqrt{3}, x_{4}=2-\sqrt{3}$.