Let's determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that

$$
(x+1) \cdot f(x+2)-2(x+2) \cdot f(-x-1)=3 x^{2}+8 x+3
$$
Solution. If $a$ is any real number, then the value of $f(a)$ is only determined by $f(1-a)$, since $(x+2)+(-x-1)=1$. Take any number $y$. Substitute $x=y$ into the functional equation:

$$
(y+1) \cdot f(y+2)-2(y+2) \cdot f(-y-1)=3 y^{2}+8 y+3
$$

With the same $y$, let $x=-y-3$:

$$
\begin{aligned}
(-y-2) \cdot f(-y-1)-2(-y-1) \cdot f(y+2) & =3(-y-3)^{2}+8(-y-3)+3 \\
2(y+1) \cdot f(y+2)-(y+2) \cdot f(-y-1) & =3 y^{2}+18 y+27-8 y-24+3 \\
4(y+1) \cdot f(y+2)-2(y+2) \cdot f(-y-1) & =6 y^{2}+20 y+12
\end{aligned}
$$

Subtract the first equation from this:

$$
3(y+1) \cdot f(y+2)=3 y^{2}+12 y+9
$$

that is,

$$
3(y+1) \cdot f(y+2)=3(y+1)(y+3)
$$

If $y \neq-1$, we can divide by $3(y+1)$:

$$
f(y+2)=y+3
$$

We have found that if $y \neq-1$, or $x \neq 1$, then $f(x)=x+1$.

The value of $f(1)$ can only be determined if we choose $x$ to be -1 or -2 in the original functional equation. However, in both cases, the coefficient of $f(1)$ is 0, so its value cannot be determined from the equation. Therefore,

$$
f(x)= \begin{cases}x+1, & \text { if } x \neq 1 \\ \text { arbitrary, } & \text { if } x=1\end{cases}
$$