Example 5 Find all real-coefficient polynomials $P(x)$ that satisfy the following relation,
$$
\left(x^{3}+3 x^{2}+3 x+2\right) \cdot P(x-1)=\left(x^{3}-3 x^{2}+3 x-2\right) \cdot P(x) \text {, }
$$
for all real numbers $x$.
(2003 Vietnam Olympiad)
From $\left(x^{3}+3 x^{2}+3 x+2\right) \cdot P(x-1)=\left(x^{3}-3 x^{2}+3 x-2\right) \cdot P(x)$, we know that $(x+2)\left(x^{2}+x+1\right) \cdot P(x-1)=(x-2)\left(x^{2}-x+1\right) \cdot P(x)$.
(1)
Substituting $x=-2$ into (2), we get $0=-28 \cdot P(-2)$, which implies $P(-2)=0$;
Substituting $x=2$ into (2), we get $28 \cdot P(1)=0$, which implies $P(1)=0$;
Substituting $x=-1$ into (2), we get $0=-9 \cdot P(-1)$, which implies $P(-1)=0$;
Substituting $x=1$ into (2), we get $9 \cdot P(0)=0$, which implies $P(0)=0$.
Thus, we can let $P(x)=(x-1) x(x+1)(x+2) \cdot Q(x)$,
where $Q(x)$ is a polynomial with real coefficients.
Therefore, $P(x-1)=(x-2)(x-1) x(x+1) \cdot Q(x-1)$.
Using (1), (2), (3), (1), we get
$$
\begin{array}{l}
(x-2)(x-1) \cdot x \cdot(x+1)(x+2) \cdot\left(x^{2}+x+1\right) \cdot Q(x-1) \\
=(x-2)(x-1) x(x+1)(x+2)\left(x^{2}-x+1\right) \cdot Q(x) .
\end{array}
$$

Since both sides of the above equation are polynomials in $x$, then
$$
\left(x^{2}+x+1\right) \cdot Q(x-1)=\left(x^{2}-x+1\right) \cdot Q(x) \text {. }
$$

Noting that $x^{2}+x+1$ and $x^{2}-x+1$ are coprime, we have $Q(x)=\left(x^{2}+x+1\right) \cdot R(x)$,
where $R(x)$ is a polynomial with real coefficients.
At the same time, we get $Q(x-1)=\left(x^{2}-x+1\right) \cdot R(x-1)$.
Using (5), (5), (7), we get
$\left(x^{2}+x+1\right)\left(x^{2}-x+1\right) \cdot R(x-1)=\left(x^{2}-x+1\right) \cdot\left(x^{2}+x+1\right) \cdot R(x)$. Since $\left(x^{2}+x+1\right)\left(x^{2}-x+1\right) \neq 0$, we have $R(x-1)=R(x)$, thus $R(x)$ is a constant.
From (3), (6), we get $P(x)=c(x-1) x(x+1)(x+2)\left(x^{2}+x+1\right)$, where $c$ is an arbitrary constant. Direct verification shows that the above polynomial satisfies the given conditions, and thus is the required polynomial.