Double Sum with a Neat Result Based on an interesting question here (second question), I have devised a similar one. 
Evaluate the following double sum without expansion and substitution of standard sum-of-integers formula. 
$$\sum_{x=1}^n\sum_{y=1}^n (n-x+y)$$
 A: $$\sum_{x=1}^n\sum_{y=1}^n (n-x+y)$$
$$=\sum_{x=1}^n\left(\sum_{y=1}^nn-\sum_{y=1}^nx+\sum_{y=1}^ny\right)$$
$$=\sum_{x=1}^n\left(n\cdot n-n\cdot x+\sum_{y=1}^ny\right)$$
$$=\sum_{x=1}^nn\cdot n-\sum_{x=1}^nn\cdot x+\sum_{x=1}^n\sum_{y=1}^ny$$
$$=n\cdot n\cdot n-n\cdot\sum_{x=1}^n x+n\sum_{y=1}^ny$$
$$=n\cdot n\cdot n\underbrace{-n\cdot\sum_{k=1}^n k+n\sum_{k=1}^nk}_\text{Change in index does not change the scenario}$$
$$=n^3$$
Will this do? Or is this an expansion?
A: By symmetry, the double sums on $x$ and $y$ are equal and cancel, hence $n^3$.

Shorter but a little more obfuscated:
$$\bar n+\bar x-\bar y=n\to n^3.$$
A: Here's a slightly different way to look at it. First, we rewrite the sum as:
$$\sum_{x=1}^{n} \sum_{y=1}^{n} (n-x+y) = \sum_{(x,y)\in\{1,\dots,n\}^2} (n-x+y) \enspace,$$
with the usual meaning of $A^2$ as the set of all ordered pairs of elements of the set $A$. Then, we observe that for $x\neq y$, the sum of the two terms corresponding to the pairs $(x,y)$ and $(y,x)$ is:
$$(n-x+y) + (n-y+x) = 2n$$
Therefore, we can think of each of the two pairs as contributing $n$ to the sum. For $x=y$, the term corresponding to $(x,y)$ is just $n$.
Conclusion: every pair contributes $n$ to the sum, so the sum is $n^2\cdot n= n^3$.
A: $$\begin{align}
S&=\sum_{x=1}^n\sum_{y=1}^n (n-x+y)\\
&=\sum_{r=1}^n\sum_{s=1}^n (n+r-s)
&&\scriptsize (r=n+1-x, s=n+1-y)\\
&=\sum_{x=1}^n\sum_{y=1}^n(n+x-y)
&&\scriptsize\text{(replacing $r,s$ with $x,y$, or directly, by symmetry)}\\
2S&=\sum_{x=1}^n \sum_{y=1}^n (n-x+y)+(n+x-y)\\
&=2\sum_{x=1}^n\sum_{y=1}^n n\\
S&=\sum_{x=1}^n\sum_{y=1}^n n\\
&=\color{red}{n^3}\end{align}$$
A: Last two sums are equal in the first line:$$\begin{array}{}\displaystyle\sum_{x=1}^n\sum_{y=1}^n(n-x+y)&\displaystyle\vphantom{\cfrac11}=\sum_{x=1}^n\sum_{y=1}^nn&\displaystyle-\sum_{x=1}^n\sum_{y=1}^nx&\displaystyle+\sum_{x=1}^n\sum_{y=1}^ny\\\vphantom{\cfrac11}&\displaystyle=n\sum_{x=1}^n1\sum_{y=1}^n1\\\vphantom{\cfrac11}&\displaystyle=n\cdot n\cdot n\\&=n^3\end{array}$$
