<s>[INST] Simplification of $\sqrt{14} - \sqrt{16 - 4 \sqrt{7}}$ I was trying to simplify $\sqrt{14} - \sqrt{16 - 4 \sqrt{7}}$. Numerical evaluation suggested that the answer is $\sqrt{2}$ and it checked out when I substituted $\sqrt{2}$ in the equation $x= \sqrt{14} - \sqrt{16 - 4 \sqrt{7}}$.
But I still cannot simplify the initial expression by any obvious means (squaring, multiplying by conjugate etc). Any idea how to do it? Can it be done?
 [/INST] Suppose that
$$
\left(\sqrt{a}\pm\sqrt{b}\right)^2
=\overbrace{a+b\vphantom{\sqrt4}}^c\pm\overbrace{\sqrt{4ab}}^{\sqrt{d}}
$$
Then
$$
\begin{align}
c^2-d
&=a^2+2ab+b^2-4ab\\
&=(a-b)^2
\end{align}
$$
Thus,
$$
a=\frac{c+\sqrt{c^2-d}}2
\qquad\text{and}\qquad
b=\frac{c-\sqrt{c^2-d}}2
$$

If $c=4$ and $d=7$, we get
$$
4-\sqrt{7}=\left(\sqrt{\frac72}-\sqrt{\frac12}\right)^2
$$
Therefore,
$$
\begin{align}
\sqrt{14}-\sqrt{16-4\sqrt7}
&=\sqrt{14}-2\sqrt{4-\sqrt7}\\
&=\sqrt{14}-\left(\sqrt{14}-\sqrt2\right)\\
&=\sqrt2
\end{align}
$$
</s>