Simplifying $\sqrt[4]{161-72 \sqrt{5}}$

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

$$w = v^2\,\Rightarrow\, w' = v'^2\Rightarrow\, ww' = (vv')^2\in\Bbb Q^2$$

$$\sqrt[4]{161-72\sqrt5}=\sqrt[4]{81-72\sqrt5+80}=\sqrt[4]{(9-4\sqrt{5})^2}=\sqrt{9-4\sqrt{5}}=\sqrt{4-4\sqrt{5}+5}=\sqrt{(2-\sqrt{5})^2}=\sqrt5-2$$ The trick is to notice that $72$ factors into $2*9*4$ and since $9^2+(4\sqrt5)^2=161$ you get this