3-cocycle in $\mathbb{Z}/n\mathbb{Z}$ It is well known that $3$-cocycles in $H^3(\mathbb{Z}/n\mathbb{Z};U(1))$ are of the form
$$w(a,b,c)=\exp(\frac{2\pi ik}{n^2}a(b+c-[b+c])),$$
for $k\in\mathbb{Z}_n$, $a,b,c\in\mathbb{Z}_n$ and $[b+c]=b+c\mod{n}$.
(See https://mathoverflow.net/questions/121065/explicit-3-cocycle-of-a-cyclic-group)

How do I show that this is indeed a $3$-cocycle?

According to me, the identity that must be shown is $(\delta^3w)(a,b,c,d)=0$.
Here I use $\delta^nf(x_1,...,x_{n+1})=f(x_2,...,x_{n+1})(\Pi_{i=1}^nf(x_1,...,x_ix_{i+1},...,x_{n+1})^{(-1)^i})f(x_1,...,x_n)^{(-1)^{n+1}}$.
Since we have a product of exponentials, we can only regard the sum of the terms between brackets. Thus we will only look at $w'(a,b,c)=a(b+c-[b+c])$ and we want $(\delta^3w')(a,b,c,d)=0\mod{n^2}$.
We then obtain
$$\begin{align}
(\delta^3w')(a,b,c,d)=&w'(b,c,d)w'(a+b,c,d)^{-1}w'(a,b+c,d)w'(a,b,c+d)^{-1}w'(a,b,c)\\
\end{align}$$
$$\begin{align}
=b&(c+d-[c+d])+\\
-(a+b)&(c+d-[c+d])+\\
a&(b+c+d-[b+c+d])+\\
-a&(b+c+d-[b+c+d])+\\
a&(b+c-[b+c])\\
=a&(b-d+[c+d]-[b+c]).
\end{align}$$

How is $a(b-d+[c+d]-[b+c])$ equal to $0\mod{n^2}$?

 A: I just came across this question.  Using Roland's hint about looking at the article on arxiv, I found that the correct formula is in prop 2.3
$$w(a,b,c)=exp(\frac{2πik}{n} a[\frac{b+c}{n}]),$$ where $[x]$ denotes the greater integer less than or equal to x. It is very fast to verify this is a 3-cocycle using the trick in prop 2.1 in the braided monoidal gr-categories article. 
A: The mistake in the original question is the following: $w'(a+b, c, d) = [a + b]\cdot (c + d - [c + d])$ instead of $(a + b)\cdot (c + d - [c + d])$. So, you should make each argument modulo $n$ before using the formula of the 3-cocycle. With this changes the formula is correct. Indeed:
\begin{multline}
w'(b, c, d) - w'(a+b, c, d) + w'(a, b+c, d) - w'(a, b, c+d) + w'(a, b, c) = \\ = b\cdot(c + d - [c+d]) - [a+b]\cdot(c+d-[c+d]) + a\cdot([b+c] + d - [b+c+d]) - \\ - a\cdot(b + [c+d] - [b+c+d]) + a(b+c - [b+c]) = \\ = (a+b)\cdot (c+d) + [a+b]\cdot [c+d] - (a + b)\cdot [c + d] - [a + b]\cdot (c + d)
\end{multline}
This is quite symmetric expression. Let $a+b = \delta_{12} + [a+b]$ and $c+d = \delta_{34} + [c + d]$. Here $\delta_{12}, \delta_{34}\in\{0, n\}$, because the value $a + b$ can be less than $n$ (and in this case $\delta_{12} = 0$) or greater than $n$ (and in this case $\delta_{12} = n$). The situation with $c + d$ is the similar. Finally:
\begin{multline}
\delta^3(w')(a, b, c, d) = \delta_{12}\cdot \delta_{34} + \delta_{12}\cdot[c+d]+[a+b]\cdot \delta_{34} + [a+b]\cdot[c+d] - \\ - \delta_{12}\cdot[c+d] - [a+b]\cdot[c+d] - \delta_{34}\cdot[a+b] - [c+d]\cdot[a+b] + \\ + [a+b]\cdot[c+d] = \delta_{12}\cdot\delta_{34} = 0 (mod\ n^2).
\end{multline}
