Example 2-41 Find $S_{n}=1^{3}+2^{3}+\cdots+n^{3}$.
Solve $\Delta S_{n}=S_{n+1}-S_{n}=(n+1)^{3}$ is a 3rd degree polynomial of $n$, therefore $S_{n}$ satisfies the recurrence relation:

Set
$$
\begin{array}{l}
S_{n}-5 S_{n-1}+10 S_{n-2}-10 S_{n-3}+5 S_{n-4}-S_{n-5}=0 \\
S_{n}=A_{1}\binom{n}{1}+A_{2}\binom{n}{2}+A_{3}\binom{n}{3}+A_{4}\binom{n}{4} \\
S_{1}=1=A_{1} \\
S_{2}=1^{3}+2^{3}=9=1 \cdot\binom{2}{1}+A_{2}, A_{2}=7 \\
S_{3}=9+3^{3}=36=1 \cdot 3+7 \cdot\binom{3}{2}+A_{3}, A_{3}=12 \\
S_{4}=36+4^{3}=100=1 \cdot 4+7 \cdot 6+12 \cdot 4+A_{4}, \quad A_{4}=6
\end{array}
$$

So $S_{n}=\binom{n}{1}+7\binom{n}{2}+12\binom{n}{3}+6\binom{n}{4}$. Let's verify the result above with $n=5$ as an example.
$$
\begin{array}{c}
S_{5}=100+5^{3}=225 \\
\binom{5}{1}+7\binom{5}{2}+12\binom{5}{3}+6\binom{5}{4}=5+70+120+30=225
\end{array}
$$