I am having problems understanding the identity (more specifically the last equality) $$S_{xx} = \sum_{1}^{n}(x_i - \bar{x})^2 = \sum_{1}^{n} x_i^2 - \frac{(\sum_{1}^{n} x_i)^2}{n}. $$
I've made an effort to show it but I get stuck with
$$ \sum_{1}^{n}(x_i - \bar{x})^2 = \sum_{1}^{n} x_i^2 - \frac{2}{n} \sum_{1}^{n} x_i^2 + \frac{1}{n^2} \sum_{1}^{n} x_i^2, $$
which probably means I've done something wrong somewhere.
Furthermore, I wonder if it is true that $$ \frac{1}{n} \left( \sum_{1}^{n} x_i^2 - n\bar{x}^2 \right) = \frac{1}{n}S_{xx},$$ and if so, how do I "see" that?
Not knowing where exactly you are messing up I've posted the whole solution, $$S_{xx}=\sum_{i=1}^{n} (x_i-\bar{x})^2 = \sum_{i=1}^{n} (x_i^2-2x\bar{x}+\bar{x}^2)=\sum_{i=1}^{n} x_i^2 -2 \bar{x}\sum_{i=1}^{n} x_i +\bar{x}^2 \sum_{i=1}^{n} 1\\= \sum_{i=1}^{n} x_i^2 -2\frac{\sum_{i=1}^{n} x_i}{n}\sum_{i=1}^{n} x_i +\left(\frac{\sum_{i=1}^{n} x_i}{n}\right)^2 n \\= \sum_{i=1}^{n} x_i^2 - \frac{(\sum_{i=1}^{n} x_i)^2}{n} $$
$$\frac{1}{n}S_{xx}=\frac{1}{n}\left(\sum_{i=1}^n x_i^2- \frac{(\sum_{i=1}^{n} x_i)^2}{n}\right)\\=\frac{1}{n}\left(\sum_{i=1}^n x_i^2- n\left(\frac{\sum_{i=1}^{n} x_i)^2}{n}\right)^2\right)\\=\frac{1}{n}(\sum_{i=1}^n x_i^2 - n \bar{x}^2)$$