<s>[INST] How would you show that $\sqrt{14+4\sqrt{10}} - \sqrt{14-4\sqrt{10}} = 4$? I've recently seen a Highschool problem and I was wondering, how would you show that
$$\sqrt{14+4\sqrt{10}} - \sqrt{14-4\sqrt{10}} = 4$$
Thank you for your time,
 [/INST] $$\begin{align}\sqrt{14+4\sqrt{10}} - \sqrt{14-4\sqrt{10}} &= \sqrt{\left(\sqrt{14+4\sqrt{10}} - \sqrt{14-4\sqrt{10}}\right)^2} \\
&= \sqrt{14 + 4\sqrt{10} + 14 - 4\sqrt{10} - 2\sqrt{(14 + 4\sqrt{10})(14-4\sqrt{10})}} \\
&= \sqrt{28 - 2\sqrt{14^2 - 4^2\cdot 10}} \\
&= \sqrt{28 - 2\cdot \sqrt{196 - 160}} \\
&= \sqrt{28 - 2\sqrt{36}} \\
&= \sqrt{28 - 12} \\
&= \sqrt{16} \\
&= 4\end{align}$$ 
In the above, you're applying $(a-b)^2 = a^2 + b^2 - 2ab$ and $(a+b)(a-b) = a^2 - b^2$.
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