How to calculate the variance of $(X_1-X_2)^2$? $X_1$ and $X_2$ are both a random sample from $\mathrm{N}(\mu,\sigma^{2})$.
How to calculate variance of $(X_1-X_2)^2$?
 A: Here is the outline provided by @MarkLStone.
First, the distribution of $Z=(X_1-X_2)$ is $\mathrm{N}(0,2\sigma^{2})$.
Next, we know that a squared normal distribution with mean zero and unit variance follows a Chi-square distribution with one degree of freedom. Consequently:
$$
\frac{Z^{2}}{2\sigma^{2}}\sim \chi^{2}(\nu=1)
$$
Hence
$$
Z^{2}\sim 2\sigma^{2}\chi^{2}(\nu=1)
$$
Edit: Thanks to @Glen, there is even a faster way than that outlined below:
The variance of the Chi-squared distribution is $2\nu$. Furthermore, a basic property of the variance is $\mathrm{Var}(kX)=k^{2}\mathrm{Var}(X)$. So with $k=2\sigma^{2}$ we arrive immediately at
$$
\mathrm{Var}(Z^{2})=\mathrm{Var}(2\sigma^{2}\chi^{2}_1))=4\sigma^{4}\mathrm{Var}(\chi^{2}_1))=4\sigma^{4}\cdot2=8\sigma^{4}
$$

Old version
The next step is to know the following relationship between a Chi-squared distribution and the gamma distribution:
$$
\mathrm{If}\: X\sim\chi^{2}(\nu)\;\mathrm{and}\;c>0, \;\mathrm{then}\; cX\sim\mathrm{Gamma}\left(k=\nu/2, \theta=2c\right)
$$
So we have at last:
$$
Z^{2}\sim \mathrm{Gamma}\left(k=1/2, \theta=2\left(2\sigma^{2}\right)\right)
$$
The variance of a gamma distribution is $k\theta^{2}$. So we end up with
$$
\mathrm{Var}(Z^{2})=\frac{1}{2}\cdot \left(4\sigma^{2}\right)^{2}=8\sigma^{4}
$$
A: Denote
\begin{eqnarray*}
z &=&\left( X_{1}-X_{2}\right) ^{2} \\
&=&X_{1}^{2}-2X_{1}X_{2}+X_{2}^{2}
\end{eqnarray*}
Therefore, 
$$
\mathbb{E}\left( z\right) =2\sigma ^{2}
$$
and 
\begin{eqnarray*}
z^{2} &=&\left( X_{1}-X_{2}\right) ^{4} \\
&=&X_{1}^{4}-4X_{1}^{3}X_{2}+6X_{1}^{2}X_{2}^{2}-4X_{1}X_{2}^{3}+X_{2}^{4}
\end{eqnarray*}
now all you need to remember is that if $X_{1}\sim \mathcal{N}\left( \mu
,\sigma ^{2}\right) $ then 
$
\mathbb{E}\left( X_{1}^{3}\right) =\mu ^{3}+3\mu \sigma ^{2}
$
and $\mathbb{E}\left( X_{1}^{4}\right) =\mu ^{4}+6\mu ^{2}\sigma ^{2}+3\sigma ^{4}
$
similarly to $X_{2}$ hence
\begin{eqnarray*}
\mathbb{E}\left( z^{2}\right)  &=&2\left( \mu ^{4}+6\mu ^{2}\sigma
^{2}+3\sigma ^{4}\right) -8\mu \left( \mu ^{3}+3\mu \sigma ^{2}\right)
+6\left( \mu ^{2}+\sigma ^{2}\right) ^{2} \\
&=&12\sigma ^{4}
\end{eqnarray*}%
\begin{eqnarray*}
Var\left( z\right)  &=&\mathbb{E}\left( z^{2}\right) -\left( \mathbb{E}%
\left( z\right) \right) ^{2} \\
&=&12\sigma ^{4}-4\sigma ^{4} \\
&=&8\sigma ^{4}
\end{eqnarray*}
