468. Solve the equation:

$$
x^{6}+3 x^{5}+2 x^{4}-2 x^{2}-3 x-1=0
$$
468. It is not difficult to see that the equation is satisfied by the value $x_{1}=1$.

Therefore, dividing by $x-1$, we get:

$$
\begin{gathered}
x^{5}+4 x^{4}+6 x^{3}+6 x^{2}+4 x+1=0 \\
\left(x^{5}+1\right)+4 x^{3}(x+1)+6 x^{2}(x+1)=0
\end{gathered}
$$

from which

$$
x_{2}=-1
$$

and

$$
x^{4}+3 x^{3}+3 x^{2}+3 x+1=0
$$

This is already a standard reciprocal equation of the 4th degree. Dividing by $x^{2}$ and making the substitution $x+\frac{1}{x}=z$, we can easily find:

$$
x_{3,4}=\frac{-6+2 \sqrt{5} \pm \sqrt{-2-6 \sqrt{5}}}{4}
$$

200

$$
x_{5,6}=\frac{-6-2 \sqrt{5} \pm \sqrt{-2+6 \sqrt{5}}}{4}
$$