Find integers $a, b$, and $c$ such that
$$
m^{3}=a\left(\begin{array}{c}
m \\
3
\end{array}\right)+b\left(\begin{array}{c}
m \\
2
\end{array}\right)+c\left(\begin{array}{c}
m \\
1
\end{array}\right)
$$
for all $m$. Then sum the series $1^{3}+2^{3}+3^{3}+\cdots+n^{3}$.