Factoring the likelihood function of a sample $X_1, ...,X_n$ where $X_i \sim N(\mu ,\sigma^2 )$ and iid to include $\bar{x} $

Question

I'm having trouble factoring the exponent of the likelihood function to include $\bar{x} $ so that I may calculate the likelihood ratio statistic.

In short, I need to show that $$\sum_{i=1}^n (x_i-\mu)^2=\sum_{i=1}^n(x_i-\bar{x})^2+\sum_{i=1}^n(\bar{x}-\mu)^2$$

For some reason (maybe lack of sleep), I can't see it.

Answer 1

This follows from $$\sum(x_i-\mu)^2=\sum(x_i-\bar x + \bar x -\mu)^2=\sum(x_i-\bar x)^2 + \sum(\bar x -\mu)^2+2(\bar x -\mu)\sum(x_i-\bar x)$$ and $$\sum(x_i-\bar x)=\sum x_i-N\bar x=\sum x_i-N(\frac{1}{N}\sum x_i)=0$$