How we can show this: $\sqrt{3+\sqrt{3}}-\sqrt{3-\sqrt{3}}=\sqrt{6-2\sqrt{6}}$ Only way in my mind to show that : $$\sqrt{3+\sqrt{3}}-\sqrt{3-\sqrt{3}}=\sqrt{6-2\sqrt{6}}$$   is to multiply $\sqrt{3+\sqrt{3}}-\sqrt{3-\sqrt{3}}$ by the conjugate factor which is $\sqrt{3+\sqrt{3}}+\sqrt{3-\sqrt{3}}$ such that we can get :$\frac{2\sqrt{3}}{\sqrt{3+\sqrt{3}}+\sqrt{3-\sqrt{3}}}$ then no simplification I can do after that , any way ?

Note that $3 + \sqrt{3} > 0$ obviously and thus $\sqrt{3+\sqrt{3}} >0$. Again, obviously $\sqrt{3-\sqrt{3}} < \sqrt{3+\sqrt{3}}$ and thus the LHS is positive.
Also, $6-2\sqrt{6} >0 \Leftrightarrow 3 > \sqrt{6}$ which holds, as $3 = \sqrt{9} > \sqrt{6}$ as $f(x) = \sqrt{x}$ is a strictly increasing function.
This is important to show for a rigorous proof, as proceeding by squaring in another case would be wrong.
Now :
$$\sqrt{3+\sqrt{3}}-\sqrt{3-\sqrt{3}}=\sqrt{6-2\sqrt{6}} \implies \bigg(\sqrt{3+\sqrt{3}}-\sqrt{3-\sqrt{3}}\bigg)^2=\bigg(\sqrt{6-2\sqrt{6}}\bigg)^2$$
Can you now apply the standard formula $(a+b)^2 = a^2 + 2ab + b^2$ to show that the desired equality formula indeed holds ?
