4. If the function $f(x)$ satisfies
$$
\begin{array}{l}
f(x+y)=f(x)+f(y)+xy(x+y) \\
f^{\prime}(0)=1 .
\end{array}
$$

Find the analytical expression of the function $f(x)$.
提示:由原式得
$$
\begin{array}{l}
f(x+y) \\
=f(x)+f(y)+\frac{1}{3}\left[(x+y)^{3}-x^{3}-y^{3}\right] .
\end{array}
$$

上式可化为柯西方程
$$
g(x+y)=g(x)+g(y) \text {, }
$$

其中, $g(x)=f(x)-\frac{1}{3} x^{3}$.
则 $f(x)-\frac{1}{3} x^{3}=a x$.
再由 $f^{\prime}(0)=1$, 得 $a=1$.
于是, $f(x)=\frac{1}{3} x^{3}+x$.

Hint: From the original equation, we get
$$
\begin{array}{l}
f(x+y) \\
=f(x)+f(y)+\frac{1}{3}\left[(x+y)^{3}-x^{3}-y^{3}\right] .
\end{array}
$$

The above equation can be transformed into Cauchy's equation
$$
g(x+y)=g(x)+g(y) \text {, }
$$

where, $g(x)=f(x)-\frac{1}{3} x^{3}$.
Then, $f(x)-\frac{1}{3} x^{3}=a x$.
By $f^{\prime}(0)=1$, we get $a=1$.
Thus, $f(x)=\frac{1}{3} x^{3}+x$.