2. Find all real roots of the equation

$(x+1)^{5}+(x+1)^{4}(x-1)+(x+1)^{3}(x-1)^{2}+(x+1)^{2}(x-1)^{3}+(x+1)(x-1)^{4}+(x-1)^{5}=0$
Solution. Multiply both sides of the equation by $(x+1)-(x-1)$ (this factor equals 2). Use the formula $a^{6}-b^{6}=(a-b)\left(a^{5}+a^{4} b+a^{3} b^{2}+a^{2} b^{3}+a b^{4}+b^{5}\right)$.

$$
\begin{gathered}
(x+1)^{6}-(x-1)^{6}=0 \\
(x+1)^{6}=(x-1)^{6}
\end{gathered}
$$

The equation

$$
x+1=x-1
$$

has no solutions. The equation

$$
x+1=-(x-1)
$$

has the solution $x=0$. Since we multiplied the equation by the number 2, we did not obtain any extraneous roots.

Answer. 0.