If $x^{2}-2 \sqrt{10} x+1=0$, what is the value of $\left(x-\frac{1}{x}\right) ?$

$x^{2}-2 \sqrt{10} x+1=0$

$\Rightarrow x^{2}+1=2 \sqrt{10} x$

$\Rightarrow \frac{x^{2}+1}{x}=2 \sqrt{10}$

$\Rightarrow x+\frac{1}{x}=2 \sqrt{10}$

On squaring both sides,

$x^{2}+\frac{1}{x^{2}}+2=(2 \sqrt{10})^{2}=40$

$\Rightarrow x^{2}+\frac{1}{x^{2}}=40-2=38$

$\Rightarrow\left(x-\frac{1}{x}\right)^{2}+2=38$

$\Rightarrow\left(x-\frac{1}{x}\right)^{2}=38-2=36$

$\therefore x-\frac{1}{x}=\sqrt{36}=6$