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?
 A: A (more complicated) approach that works on any nested radical, would be to use the Zippel Denesting Theorem.
$\sqrt[4]{161-72\sqrt{5}}$ is a fourth power exponent in $\mathbb{Q}(\sqrt{5})$ so setting the radical equal to its primitive root of unity and finding its roots gives us the simplification.
So we have: $\sqrt[4]{161-72\sqrt{5}}=x\iff x^4+72\sqrt{5}-161=0\iff (\sqrt{5}-2-x)(x+\sqrt{5}-2)(4\sqrt{5}-9-x^2)=0$ with the first one giving the correct denesting of $\sqrt{5}-2$.
A: Denesting $\sqrt w = \sqrt{a+b\sqrt{n}}\,$ can be done by a simple formula that I discovered in my youth.
$ {\bf Simple\ Denesting\ Rule} \ \ \overbrace{\rm \color{#0a0}{subtract\ out}\ \sqrt{norm}^{\phantom .}}^{\textstyle\!\!\! w \to w - \sqrt{ww'} =:\, s\!\!\!\!\!\!}\!\!\!, \ {\rm then}\ \  \overbrace{\color{brown}{\rm divide\ out}\ \sqrt{{\rm trace}}^{\phantom .}}^{\textstyle s\,\to\, s/\sqrt{s+s'}\!\!\!\!\!\!\!}$ from that.
$\!\begin{align}{\rm Recall}\ \ w = a + b\sqrt{n}\rm \ \ has\ \ {\bf norm}\  &=\: w\:\cdot\: w' = (a + b\sqrt{n})\ \cdot\: (a - b\sqrt{n})\ =\: a^2\! - n\: b^2\\[4pt]
{\rm and,\ furthermore,\ }w\rm \ \ has\ \ {\bf trace}\ &=\: w+w' =  (a + b\sqrt{n}) + (a - b\sqrt{n})\: =\:  2\,a\end{align}$
In the  norm/trace sqrts either sign works e.g. $\sqrt 1 = \pm1,\,$ so we choose what proves simplest.

Here $\:161-72\sqrt 5\:$ has norm $= 1.\:$ $\rm\ \color{#0a0}{subtracting\ out}\ \sqrt{norm}\ = -1\ $ yields $\  162-72\sqrt 5\:$
which has $\ {\rm\ \sqrt{trace}}\: =\: \sqrt{324}\ =\ 18.\ \ \ \ \rm \color{brown}{Dividing\ it\ out}\ \,$ of the above  yields $\ \ \ 9-4\sqrt 5$
Checking: $\ (9 - 4\sqrt 5)^2 = 9^2\!+\! 4^2(5)- 2(9)4\sqrt 5 = 161-72\sqrt 5 \ \ \large \color{#c00}\checkmark$

Next $\:9-4\sqrt 5\:$ has norm $= 1.\:$ $\rm\ \color{#0a0}{subtracting\ out}\ \sqrt{norm}\ = 1\ $ yields $\  8-4\sqrt 5\:$
with $\ {\rm\ \sqrt{trace}}\: =\: \sqrt{16}\ =\ 4.\ \ \ \ \rm \color{brown}{Dividing\ it\ out}\,\ $ of the above  yields $\,\ \ \ 2-\sqrt 5$
Checking: $\ (2 - \sqrt 5)^2 = 2^2\!+\! 5 - 2\cdot 2\sqrt 5 = 9 - 4\sqrt 5 \ \ \large \color{#c00}\checkmark$
Negating $\,2-\sqrt 5\,$ to get the positive square-root yields the sought result. We chose the signs in  $\,\sqrt 1 = \pm 1$ so that  arithmetic is simplest. Any choice will work as the proof below shows (e.g. we do both here). For many worked examples see prior posts on denesting.  Below is a sketch of a proof.
Lemma $\ \  \sqrt w\, =\, \dfrac{s}t,\ \ \ \begin{align}s &\,=\, w \pm \sqrt{ww'}\\[.1em] t &\,=\: \pm\sqrt{s+s'}\end{align}\ $ when $\ \ \color{#90f}{\sqrt{ww'}\in\Bbb Q}$
Proof $\quad\ s^2 =\, w (w+w' \pm 2\sqrt{ww'})\, =\, w\, t^2$
Necessarily $\ \color{#90f}{\sqrt{ww'}\in \Bbb Q}\,$ if a denesting $\sqrt w = v = c + d\sqrt n\,$ exists, since
$$w = v^2\,\Rightarrow\, w' = v'^2\Rightarrow\, ww' = (vv')^2\in\Bbb Q^2$$
A: I think this problem is a lot easier than the other answers would have one believe. Since
$$161^2-5\cdot72^2=(161+72\sqrt5)(161-72\sqrt5)=1$$
We can see that $161+72\sqrt5$ is a unit in the ring of integers of $\mathbb{Q}(\sqrt5)$ Thus it must be $(\pm1)$ times a power of the fundamental unit, $\phi=\frac{1+\sqrt5}2$:
$$161+72\sqrt5=\phi^n$$
Solving for $n$ we have
$$n=\frac{\ln(161+72\sqrt5)}{\ln\left(\frac{1+\sqrt5}2\right)}=12$$
Thus
$$(161+72\sqrt5)^{1/4}=\phi^3=2+\sqrt5$$
A: $$\sqrt[4]{161-72\sqrt5}=\sqrt[4]{81-72\sqrt5+80}=\sqrt[4]{(9-4\sqrt{5})^2}=\sqrt{9-4\sqrt{5}}=\sqrt{4-4\sqrt{5}+5}=\sqrt{(2-\sqrt{5})^2}=\sqrt5-2$$
The trick is to notice that $72$ factors into $2*9*4$ and since $9^2+(4\sqrt5)^2=161$ you get this
A: Number's Simple Denesting Rule (https://math.stackexchange.com/q/816527) is incorrect. According to it, 
$ \sqrt{a + b \sqrt{n}} = \frac{a + b \sqrt{n} - \sqrt{(a + b \sqrt{n})(a - b \sqrt{n})}}{\sqrt{2 a}}.$
The latter formula is equivalent to 
$\sqrt{a + b \sqrt{n}} - \sqrt{a - b \sqrt{n}} = \sqrt{2a},$
which is not an identity. 
Moreover, according to Number's definition of the trace of an expression $a + b \sqrt{n}$, the trace for $161 - 72 \sqrt{5}$ is $322$ (not $324$, as Number wrote) and the trace for $9 - 4 \sqrt{5}$ is $18$ (not $16$). However, these miscalculations together with the incorrect Simple Denesting Rule led to the correct results 
$\sqrt{161 - 72 \sqrt{5}} = 9 - 4 \sqrt{5}$, $\qquad \sqrt{9 - 4 \sqrt{5}} = 2- \sqrt{5}$. 
A: 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*}
