Given that $\sum^{n}_{r=-2}{r^3}$ can be written in the form $an^4+bn^3+cn^2+dn+e$, show that: $\sum^{n}_{r=0}{r^3}=\frac14n^2(n+1)^2$ Question:

Given that $\displaystyle\sum^{n}_{r=-2}{r^3}$ can be written in the form $an^4+bn^3+cn^2+dn+e$, show that: $$\sum^{n}_{r=0}{r^3}=\frac14n^2(n+1)^2$$

Attempt:
Substituting $n = -2,-1,0,1,2$into $\sum_{r=-1}^{n}{r^3}$ we get:
$$\sum_{r=-2}^{-2}{r^3} = -8 =16a-8b+4c-2d+e$$
$$\sum_{r=-2}^{-1}{r^3} = -9 = a-b+c-d+e$$
$$\sum_{r=-2}^{0}{r^3} = -9 = e$$
$$\sum_{r=-2}^{1}{r^3} = -8 = a+b+c+d+e$$
$$\sum_{r=-2}^{2}{r^3} = 0 =16a+8b+4c+2d+e$$
After some algebra we now know $a= -\frac12, b = -1, c = 1, d = \frac32, e = -9$:
$$\therefore \sum^{n}_{r=-2}{r^3}= -\frac12n^4 -n^3 + n^2 +\frac32n -9$$
We define $\sum^{n}_{r=0}{r^3}$:
$$\sum^{n}_{r=0}{r^3}=\sum^{n}_{r=-2}{r^3} - \sum^{-2}_{r=-2}{r^3}=-\frac12n^4 -n^3 + n^2 +\frac32n -9 -(-9)$$
$$-\frac12n^4 -n^3 + n^2 +\frac32n = \frac14(-2n^4-4n^3+4n^2 + 6n)$$
$$\frac14n^2\biggl(-2n^2-4n+4+\frac6n\biggl) = \frac14n^2\biggl(\frac{-2n^3-4n^2+4n+6}{n}\biggl)$$
$$ \frac14n^2\biggl(\frac{-2n^3-4n^2+4n+6}{n}\biggl)$$
My problem:
I got $d = \frac32$ and as far as i can see since it need to end up with $d = 0$ as the smallest term that can be left is a $n^2$ term as $\frac14n^2(n+1)^2$ is multiplied by $n^2$ meaning $e$ has to cancel out (which it does) but i don't know how to get rid of $d$, if you could explain it to me it would be much appreciated

for $n \ge k$ let the sum $\sum_{r=k}^n r^3 $ be $S_{k}(n)$ so for $n \ge 0$
$$
S_{-2}(n) = S_0(n) -9
$$
since $S_0(0)=0$ this gives $S_{-2}(0)=e=-9$ and
$$
S_0(n)= an^4+bn^3+cn^2+dn
$$
now $S_0(n)-S_0(n-1) = n^3$
so
$$
\begin{align}
a(4n^3-6n^2+4n-&1) + \\
b(3n^2-3n+&1)+\\
c(2n-&1)+\\
&d= n^3
\end{align}
$$
comparing coefficients of $n^3$ gives $a=\frac14$.
the $n^2$ term gives $b=2a$ so $b=\frac12$.
the term in $n$ gives $4a-3b+2c=0$ so $c-\frac14$. and for the constant term: $a-b+c-d=0$ so $d=0$.
altogether this gives
$$
S_0(n)=\frac14 n^4 + \frac12 n^3 + \frac14 n^2 = \bigg(\frac{n(n+1)}2 \bigg)^2
$$
