2. Solve the equation

$$
(1+x)^{8}+\left(1+x^{2}\right)^{4}=2 x^{4}
$$
Solution. Since $x=0$ is not a solution to the equation, it is equivalent to the equation

$$
\left(\frac{x^{2}+2 x+1}{x}\right)^{4}+\left(\frac{x^{2}+1}{x}\right)^{4}-2=0
$$

By the substitution, $\frac{x^{2}+x+1}{x}=p$, we obtain the equation

$$
(p-1)^{4}+(p+1)^{4}-2=0
$$

whose solutions are $p_{1}=p_{2}=0, p_{3}=i \sqrt{6}$ and $p_{4}=-i \sqrt{6}$.

For $p_{1}=p_{2}=0$, we get $x^{2}+x+1=0$, i.e., $x_{1}=x_{2}=\frac{-1+i \sqrt{3}}{2}, x_{3}=x_{4}=\frac{-1-i \sqrt{3}}{2}$.

For $p_{3}=i \sqrt{6}$, we get the equation $x^{2}+(1-i \sqrt{6}) x+1=0$, i.e.,

$$
x_{5 / 6}=\frac{-1+i \sqrt{6} \pm \sqrt{-9-2 i \sqrt{6}}}{2},
$$

For $p_{3}=-i \sqrt{6}$, we get the equation $x^{2}+(1+i \sqrt{6}) x+1=0$, i.e.,

$$
x_{7 / 8}=\frac{-1-i \sqrt{6} \pm \sqrt{2 i \sqrt{6}-9}}{2}
$$