2.1. Find all functions $f: \mathbb{R} \longrightarrow \mathbb{R}$ such for any $x, y \in \mathbb{R}$,
$$
(x-y) f(x+y)-(x+y) f(x-y)=4 x y\left(x^{2}-y^{2}\right) \text {. }
$$
2.1. 1st solution

Let $x=y=1$. We have $f(0)=0$. Let $a=x+y$ and $b=x-y$. Then the given functional equation is equivalent to $b f(a)-a f(b)=\left(a^{2}-b^{2}\right) a b$. This holds for all real numbers $a$ and $b$. For nonzero $a$ and $b$, this can be rewritten as
$$
\frac{f(a)}{a}-a^{2}=\frac{f(b)}{b}-b^{2} .
$$

Hence, for any nonzero real number $x, \frac{f(x)}{x}-x^{2}=f(1)-1$. Let $\alpha=f(1)-1$. We have $f(x)=x^{3}+\alpha x$, for all $x \neq 0$. As $f(0) \stackrel{x}{=} 0$, we thus have $f(x)=x^{3}+\alpha x$ for all $x \in \mathbb{R}$. Clearly $f(x)=x^{3}+\alpha x$ satisfies the given relation.