<s>[INST] Why $\sqrt{23-\sqrt{17}}-2\sqrt{7-\sqrt{17}}=\sqrt{71-17\sqrt{17}}$ is true? Easy to show this identity after squaring twice of the both sides.
But why it turned out true?
For example, if we want to prove that
$$\sqrt{23-3\sqrt{5}}-2\sqrt{3-\sqrt{5}}=\sqrt{3+\sqrt{5}},$$
we can do it without squaring:
$$\sqrt{23-3\sqrt{5}}-2\sqrt{3-\sqrt{5}}=\frac{1}{\sqrt{2}}\left(\sqrt{46-6\sqrt{5}}-2\sqrt{6-2\sqrt{5}}\right)=$$
$$=\frac{1}{\sqrt{2}}\left(\sqrt{(3\sqrt{5}-1)^2}-2\sqrt{(\sqrt5-1)^2}\right)=\frac{1}{\sqrt{2}}\left(3\sqrt{5}-1-2(\sqrt{5}-1)\right)=$$
$$=\frac{1}{\sqrt{2}}(\sqrt{5}+1)=\frac{1}{\sqrt{2}}\sqrt{6+2\sqrt{5}}=\sqrt{3+\sqrt{5}}.$$
But this way does not work for the starting identity. 
How to prove the starting identity without squaring? 
Thank you!  
 [/INST] Thumb rule to merge any two irrationals: $\boxed{\color{red}{\sqrt{a}\pm\sqrt{b}=\sqrt{(\sqrt{a}\pm\sqrt{b})^2}} \quad \forall \  \color{blue}{a>b}}$ 
$$\therefore \sqrt{23-\sqrt{17}}-2\sqrt{7-\sqrt{17}}$$
$$=\sqrt{\left(\sqrt{23-\sqrt{17}}-2\sqrt{7-\sqrt{17}}\right)^2}$$
$$=\sqrt{23-\sqrt{17}+4(7-\sqrt{17})-4\sqrt{(23-\sqrt{17})(7-\sqrt{17})}}$$
$$=\sqrt{23-\sqrt{17}+4(7-\sqrt{17})-4\sqrt{(23-\sqrt{17})(7-\sqrt{17})}}$$
$$=\sqrt{51-5\sqrt{17}-4\sqrt{178-30\sqrt{17}}}$$
$$=\sqrt{51-5\sqrt{17}-4\sqrt{178-2\cdot 5\cdot 3\sqrt{17}}}$$
$$=\sqrt{51-5\sqrt{17}-4\sqrt{(3\sqrt{17}-5)^2}}$$
$$=\sqrt{51-5\sqrt{17}-4(3\sqrt{17}-5)}$$
$$=\sqrt{51-5\sqrt{17}-12\sqrt{17}+20}$$
$$=\sqrt{71-17\sqrt{17}}$$
Proved.
</s>