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
 A: I would propose a more efficient strategy. Given that $\sum_{r=-2}^{n}r^3$ is a fourth-degree polynomial in the variable $n$, the same holds for $\sum_{r=0}^{n}r^3=\sum_{r=1}^{n}r^3$, since the two sums differ by a constant. Two different fourth-degree polynomials cannot agree on $5$ points or more: otherwise their difference would be a non-constant polynomial with degree $\leq 4$ and with $\geq 5$ roots, impossible. So, in order to prove
$$ \sum_{r=0}^{n}r^3 = \frac{n^2(n+1)^2}{4} \tag{1}$$
for any $n\in\mathbb{N}$, it is enough to check that such identity holds at $n\in\{0,1,2,3,4\}$:
$$ \begin{array}{|c|c|c|c|c|c|}\hline n & 0 & 1 & 2 & 3 & 4 & 5\\
\hline \sum_{r=0}^{n}r^3 & 0 & 1 & 9 & 36 & 100 & 225\\
\hline \frac{1}{4}\left(n(n+1)\right)^2 & 0 & 1 & 9 & 36 & 100 & 225 
\\\hline  \end{array}\tag{2}$$
and we are done.
A: 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
$$
S_0(n)=\frac14 n^4 + \frac12 n^3 + \frac14 n^2 = \bigg(\frac{n(n+1)}2 \bigg)^2
$$
A: $$\begin{align}
&n=0: &e&=-9\tag{1}\\\\
&n=1: &a+b+c+d&=1\tag{2}\\
&n=-1: &a-b+c-d&=0\tag{3}\\\\
\frac 12[(2)+(3)]: &&a+c&=\frac 12\tag{4}\\
\frac 12[(2)-(3)]: &&b+d&=\frac 12\tag{5}\\\\
&n=2: &16a+8b+4c+2d&=9\tag{6}\\
&n=-2: &16a-8b+4c-2d&=1\tag{7}\\\\
\frac 12 [(6)+(7)]: &&16a+4c&=5
\quad\Rightarrow a=\frac 14, c=\frac 14
\;\;\ \text{using }(4)\\
\frac 14[(6)-(7)]: &&4b+d&=2
\quad\Rightarrow b=\frac 12, d=0
\quad\text{using }(5)\\\\
&&\sum_{r=0}^n r^3&=\left(\sum_{r=-2}^n r^3\right) -9\\
&& &=an^4+bn^3+cn^2+dn+e-(-9)\\
&& &=\frac 14 n^2+\frac 12 n^3+\frac 14 n^2\\
&& &=\frac 14n^2(n+1)^2\\
&& &=\left(\frac 12 n(n+1)\right)^2\end{align}$$

See also this question I posted. 
If we can show that $n^2$ and $(n+1)^2$ are factors of the sum of odd powers $p(>1)$ of the first $n$ integers, then for $p=3$ (i.e. sum of cubes) we would just have to determine the constant ($\frac 14$, by putting $n=1$) and we are done. 
