<s>[INST] Simplifying $\sqrt[4]{161-72 \sqrt{5}}$ $$\sqrt[4]{161-72 \sqrt{5}}$$
I tried to solve this as follows:
the resultant will be in the form of $a+b\sqrt{5}$ since 5 is a prime and has no other factors other than 1 and itself. Taking this expression to the 4th power gives $a^4+4 \sqrt{5} a^3 b+30 a^2 b^2+20 \sqrt{5} a b^3+25 b^4$. The integer parts of this must be equal to $161$ and the coeffecients of the roots must add to $-72$. You thus get the simultaneous system:
$$a^4+30 a^2 b^2+25 b^4=161$$
$$4 a^3 b+20 a b^3=-72$$
In an attempt to solve this, I first tried to factor stuff and rewrite it as:
$$\left(a^2+5 b^2\right)^2+10 (a b)^2=161$$
$$4 a b \left(a^2+5 b^2\right)=-72$$
Then letting $p = a^2 + 5b^2$ and $q = ab$ you get
$$4 p q=-72$$
$$p^2+10 q^2=161$$
However, solving this yields messy roots. Am I going on the right path?
 [/INST] Another approach. We can apply twice the following general algebraic identity involving nested
radicals 
\begin{equation*}
\sqrt{a-\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^{2}-b}}{2}}-\sqrt{\frac{a-\sqrt{
a^{2}-b}}{2}}\tag{1}
\end{equation*}
to get
\begin{equation*}
\sqrt[4]{161-72\sqrt{5}}=\sqrt[4]{161-
\sqrt{25\,920}}=\sqrt{5}-2.
\end{equation*}
The numerical computation can be carried out as follows:
\begin{eqnarray*}
\sqrt[4]{161-72\sqrt{5}} &=&\left( \sqrt{\frac{161+\sqrt{161^{2}-25\,920}}{2}
}-\sqrt{\frac{161-\sqrt{161^{2}-25\,920}}{2}}\right) ^{1/2} \\
&=&\left( \sqrt{\frac{161+1}{2}}-\sqrt{\frac{161-1}{2}}\right) ^{1/2} \\
&=&\sqrt{9-\sqrt{80}} \\
&=&\sqrt{\frac{9+\sqrt{9^{2}-80}}{2}}-\sqrt{\frac{9-\sqrt{9^{2}-80}}{2}} \\
&=&\sqrt{\frac{9+1}{2}}-\sqrt{\frac{9-1}{2}}\\
&=&\sqrt{5}-2.
\end{eqnarray*}
ADDED. Note: If the radical were of the form $\sqrt{a+\sqrt{b}}$, then the applicable identity would be
\begin{equation*}
\sqrt{a+\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^{2}-b}}{2}}+\sqrt{\frac{a-\sqrt{
a^{2}-b}}{2}}.\tag{2}
\end{equation*}
Proof (from Sebastião e Silva, Silva Paulo, Compêndio de Álgebra
II, 1963). To find two rational numbers $x,y$ such that
\begin{equation*}
\sqrt{a+\sqrt{b}}=\sqrt{x}+\sqrt{y},\text{ with }a,b\in \mathbb{Q},
\end{equation*}
we square both sides and rearrange the terms
\begin{equation*}
2\sqrt{xy}=a-x-y+\sqrt{b}.
\end{equation*}
Squaring again yields
\begin{equation*}
4xy=\left( a-x-y\right) ^{2}+2\left( a-x-y\right) \sqrt{b}+b.
\end{equation*}
Since $x,y\in \mathbb{Q}$, $a-x-y=0$, which means that $x,y$ satisfy the system of equations
\begin{equation*}
x+y=a,\qquad xy=\frac{b}{4}.
\end{equation*}
Consequently they are the roots of 
\begin{equation*}
X^{2}-aX+\frac{b}{4}=0,
\end{equation*}
i.e.
\begin{eqnarray*}
x &=&X_{1}=\frac{a+\sqrt{a^{2}-b}}{2} \\
y &=&X_{2}=\frac{a-\sqrt{a^{2}-b}}{2}.
\end{eqnarray*}
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