Easiest proof of the well known normal result The well known identity that for iid normal distribution with 
$x_{i} \sim N(\mu, \sigma^{2})$
we can write
$\sum_{i=1}^{n} (x_{i}-\mu)^{2} = s^{2}+n(x-\bar x)^{2}$
where $s^{2}=\sum_{i=1}^{n}(x_{i}-\bar x)^{2}$
But it is not immediately obvious from first sight that it is true. Or is it basically just we expand out and it happens to come out.
 A: The result does not depend on the distribution of the underlying data, so the assumption of IID data and normality are unnecessary.  Regardless of the distribution of the sample values you have:
$$\begin{equation} \begin{aligned}
\sum_{i=1}^n (x_i - \bar{x}_n)
= \sum_{i=1}^n x_i - n \bar{x}_n
= n \bar{x}_n - n \bar{x}_n
= 0.
\end{aligned} \end{equation}$$
It follows that:
$$\begin{equation} \begin{aligned}
\sum_{i=1}^n (x_i - \mu)^2 
&= \sum_{i=1}^n (x_i - \bar{x}_n + \bar{x}_n - \mu)^2 \\[6pt]
&= \sum_{i=1}^n \Big[ (x_i - \bar{x}_n)^2 +2 (x_i - \bar{x}_n) (\bar{x}_n - \mu) + (\bar{x}_n - \mu)^2 \Big] \\[6pt]
&= \sum_{i=1}^n (x_i - \bar{x}_n)^2 + 2 \sum_{i=1}^n (x_i - \bar{x}_n) (\bar{x}_n - \mu) + \sum_{i=1}^n(\bar{x}_n - \mu)^2 \\[6pt]
&= \sum_{i=1}^n (x_i - \bar{x}_n)^2 + 2 (\bar{x}_n - \mu) \sum_{i=1}^n (x_i - \bar{x}_n) + n (\bar{x}_n - \mu)^2 \\[6pt]
&= \sum_{i=1}^n (x_i - \bar{x}_n)^2 + 2 (\bar{x}_n - \mu) \times 0 + n (\bar{x}_n - \mu)^2 \\[6pt]
&= \sum_{i=1}^n (x_i - \bar{x}_n)^2 + n (\bar{x}_n - \mu)^2 \\[6pt]
&= (n-1) s^2 + n (\bar{x}_n - \mu)^2. \\[6pt]
\end{aligned} \end{equation}$$
where $s^2 \equiv \sum_{i=1}^n (x_i - \bar{x}_n)^2 / (n-1)$ is the sample variance (which you have misstated in your question by removing the denominator).
