6.368 Solve the equation $x^{4}-4 x^{3}+3 x^{2}+8 x-10=0$, given that two of its roots differ only in sign.
Solution. Let $x_{1}$ and $x_{2}$ be the roots of the original equation and $x_{1}=-x_{2}$. Then
$\left\{\begin{array}{l}x_{1}^{4}-4 x_{1}^{3}+3 x_{1}^{2}+8 x_{1}-10=0, \\ x_{1}^{4}+4 x_{1}^{3}+3 x_{1}^{2}-8 x_{1}-10=0\end{array} \Rightarrow 8 x_{1}^{3}-16 x_{1}=0 \Leftrightarrow x_{1}^{2}-2=0 \Rightarrow\right.$ $x_{1}=\sqrt{2}, x_{2}=-\sqrt{2}, x^{4}-4 x^{3}+3 x^{2}+8 x-10=\left(x^{2}-2\right)\left(x^{2}-4 x+5\right) \Rightarrow$ there are only two roots: $x_{1}=\sqrt{2}, x_{2}=-\sqrt{2}$.

Answer: "roots of the equation: $x_{12}= \pm \sqrt{2}$."