Presumably you got the following system of equations:
$\left\{\begin{align*} &d=0\\ &a+b+c+d=1\\ &8a+4b+2c+d=5\\ &27a+9b+3c+d=14\;, \end{align*}\right.$
which immediately reduces to
$\left\{\begin{align*} &a+b+c=1\\ &8a+4b+2c=5\\ &27a+9b+3c=14\;. \end{align*}\right.$
Eliminate $c$ by subtracting multiples of the first equation from the other two:
$\left\{\begin{align*} &6a+2b=3\\ &24a+6b=11\;. \end{align*}\right.$
Subtracting the second equation from $4$ times the first yields $2b=1$, so $b=\frac12$, and $6a+1=3$, or $a=\frac13$. Finally, from $a+b+c=1$ we have $c=\frac16$, and the cubic polynomial is
$f(n)=\frac13n^3+\frac12n^2+\frac16n=\frac16\left(n\left(2n^2+3n+1\right)\right)=\frac16n(n+1)(2n+1)\;.$
Now you have only to prove by induction that
$\sum_{k=0}^nk^2=\frac16n(n+1)(2n+1)$
for $n\ge 0$.
Yes, this approach will work to find a closed form for $\sum_{k=0}^nk^m$ for any positive integer $a$, though it’s pretty tedious if $a$ is much bigger than $2$; $m=3$ is quite feasible, however.