Task B-2.2. (20 points) In the set of complex numbers, determine all solutions of the equation

$$
x^{8}+4 x^{6}-10 x^{4}+4 x^{2}+1=0
$$
Solution. Since 0 is not a solution, by dividing the equation by $x^{4}$ we get

$$
\begin{aligned}
x^{4}+4 x^{2}-10+\frac{4}{x^{2}}+\frac{1}{x^{4}} & =0 \\
\left(x^{4}+\frac{1}{x^{4}}\right)+4\left(x^{2}+\frac{1}{x^{2}}\right)-10 & =0
\end{aligned}
$$

By substituting $t=x^{2}+\frac{1}{x^{2}}$, we get $t^{2}=x^{4}+\frac{1}{x^{4}}+2$, so the equation becomes

$$
t^{2}+4 t-12=0
$$

whose solutions are $t_{1}=-6$ and $t_{2}=2$.

For $t_{1}=-6$ we have:

$$
\begin{array}{rl}
x^{2}+\frac{1}{x^{2}}= & -6 \\
u= & x+\frac{1}{x} \quad \text { (substitution) } \\
u^{2}= & x^{2}+2+\frac{1}{x^{2}}=-4 \\
u_{1,2}= & \pm 2 i \\
x+\frac{1}{x}=2 i & x+\frac{1}{x}=-2 i \\
x^{2}-2 i x+1=0 & x^{2}+2 i x+1=0 \\
x_{1,2}=(1 \pm \sqrt{2}) i & x_{3,4}=-(1 \pm \sqrt{2}) i
\end{array}
$$

(6 points)

For $t_{2}=2$ we get:

$$
\begin{aligned}
x^{2}+\frac{1}{x^{2}}= & 2 \\
x^{4}-2 x^{2}+1= & 0 \\
\left(x^{2}-1\right)^{2}= & 0 \\
x_{5,6}=1 & x_{7,8}=-1 .
\end{aligned}
$$

The solutions of the equation are $x \in\{ \pm 1, \pm(1 \pm \sqrt{2}) i\}$.