3A. Solve the equation

$$
\left(\frac{x^{3}+x}{3}\right)^{3}+\frac{x^{3}+x}{3}=3 x
$$
Solution. If we introduce the substitution $\frac{x^{3}+x}{3}=y$, then $y^{3}+y=3 x$ and $x^{3}+x=3 y$. Therefore, if $x_{0}$ is a solution of the equation and $\frac{x_{0}^{3}+x_{0}}{3}=y_{0}$, then $\left(x_{0}, y_{0}\right)$ is a solution of the system of equations

$$
\left\{\begin{array}{l}
x^{3}+x=y \\
y^{3}+y=x
\end{array}\right.
$$

The converse is also true, which is not difficult to prove. Therefore, it is sufficient to solve the system of equations (1).

If we subtract the second equation of the system from the first equation of the system, we get

$$
\begin{aligned}
& \left(x^{3}-y^{3}\right)+(x-y)=3(y-x) \\
& (x-y)\left(x^{2}+x y+y^{2}+1\right)=3(y-x)
\end{aligned}
$$

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

Since

$$
x^{2}+x y+y^{2}+4=x^{2}+2 x \frac{y}{2}+\frac{y^{2}}{4}+\frac{3}{4} y^{2}+4=\left(x+\frac{y}{2}\right)^{2}+\frac{3 y^{2}}{4}+4 \geq 4
$$

for any $x, y \in \mathbb{R}$, from equation (2) we have $x-y=0$ or $x=y$. Therefore $\frac{x^{3}+x}{3}=x$, from which we get $x\left(x^{2}-2\right)=0$. The solutions of the last equation are $x_{1}=0, x_{2}=\sqrt{2}, x_{3}=-\sqrt{2}$. It is not difficult to verify that these are also solutions of the original equation.