Find integers $a$, $b$, and $c$ such that $$m^3=a{m\choose3}+b{m\choose 2}+c{m\choose 1}$$ $\forall$ $m$. Then sum the series $$1^3+2^3+3^3+\cdots+n^3.$$
First, we solve for $a$, $b$, and $c$. Both sides of the equation are polynomials of degree 3. By considering the coefficient of $m^3$ on both sides, we obtain $a=6$. Evaluating both sides at $m=1$ we obtain $c=1$, and then evaluating both sides at $m=2$ we obtain $b=6$.
Where exactly do $a=6$, $b=6$, and $c=1$ come from?
$$m^3=a {m \choose 3} +b {m \choose 2}+ c{m \choose 1} = a\frac{m(m-1)(m-2)}{6}+b \frac{m(m-1)}{2} +c m.$$ $$m^3=a \frac{m^3-3m^2+2m}{6}+b\frac{m^2-m}{2}+cm=\frac{a}{3}m^3+(\frac{a}{2}-\frac{b}{2}) m^2+(\frac{a}{3}-\frac{b}{2}+c) m.$$ On comparinc thge coefficients of powers of $m$, we get $a=6, a-b=0, a/3-b/2+c=0$, then $a=6=b, c=1$, Fyrther, we use $$\sum_{k=1}^{m} {k \choose j} ={n+1 \choose j+1}.$$ to get $$S=\sum_{m=1}^{n} m^3 = \sum_{m=1}^n \left [6 {m \choose 3} +6 {m \choose 2} + 1 {m \choose 1} \right] = 6 {n+1 \choose 4} + 6 {n+1 \choose 3} + {n+1 \choose 2}$$ $$\implies S=\frac{1}{4} (n+1)n(n-1)(n-2)+(n+1)n(n-1)+\frac{1}{2}(n+1)n]=\frac{n^2(n+1)^2}{4}$$