Simple solving Skanavi book exercise: $\sqrt{9+\sqrt{80}}+\sqrt{9-\sqrt{80}}$

Simple way to solve this exercise

x = \sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}

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You are given two nested radicals. To denest each of them, we can try to find two numbers $u,v\in\mathbb {Q} $ such that

9+\sqrt{80}=\left( u+\sqrt{v}\right) ^{3},\qquad 9-\sqrt{80}=\left( u-\sqrt{v
}\right) ^{3}.\tag{1}

The product

\left( u+\sqrt{v}\right) \left( u-\sqrt{v}\right) =u^{2}-v=\sqrt[3]{9+\sqrt{

implies that



9+\sqrt{80}=\left( u+\sqrt{u^{2}-1}\right) ^{3}=4u^{3}-3u+\sqrt{\left(
u^{2}-1\right) \left( 4u^{2}-1\right) ^{2}}.\tag{2}

A solution is

4u^{3}-3u=9 \\
\left( u^{2}-1\right) \left( 4u^{2}-1\right) ^{2}=80\tag{3}


4u^{3}-3u-9 &=&4\left( u-\frac{3}{2}\right) \left( u^{2}+\frac{3}{2}u+\frac{3
( u^{2}-1) ( 4u^{2}-1) ^{2}-80
&=&16u^{6}-24u^{4}+9u^{2}-81 \\
&=&16\left( u+\frac{3}{2}\right) \left( u-\frac{3}{2}\right) \left( u^{2}-
\frac{3}{2}u+\frac{3}{2}\right) \left( u^{2}+\frac{3}{2}u+\frac{3}{2}\right)

the single solution of $(3)$ is


Consequently, $v=u^{2}-1=\frac{5}{4}$,

9+\sqrt{80}=\left( \frac{3}{2}+\frac{\sqrt{5}}{2}\right) ^{3},\qquad 9-\sqrt{
80}=\left( \frac{3}{2}-\frac{\sqrt{5}}{2}\right) ^{3}\tag{5}


x &=&\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}=\frac{3}{2}+\frac{\sqrt{5}}{2}+\frac{3}{2}-\frac{\sqrt{5}}{2}=3.\tag{6}



$$\iff x^3-3x-18=0$$

of which $x=3$ is a root(by inspection)

Find the other two roots from $$\frac{x^3-3x-18}{x-3}=0$$

Another way: notice that

If you know some algebraic number theory, the fact that $(9+\sqrt{80})(9-\sqrt{80})=1$ says that $9\pm\sqrt{80}=9\pm4\sqrt5$ is an even power of the fundamental unit ${1+\sqrt5\over2}$ in $\mathbb{Q}(\sqrt5)$, i.e. a power of $\left({1+\sqrt5\over2}\right)^2={3+\sqrt5\over2}$. It’s clear that expanding $(3+\sqrt5)^n$ will give terms that rapidly exceed $36+16\sqrt5$. In fact you can probably eyeball that $n=2$ is too small but $n=4$ is too big, so the answer must be


as observed in the answer by O.L.

For a hint:

Note that if $a+b+c=0$ then $a^3+b^3+c^3=3abc$ – can you see how to use that here?

To prove, let $a,b,c$ be the roots of $x^3-px^2+qx-r=0$ so that $p=a+b+c; q=ab+bc+ca; r=abc$ then$$a^3-pa^2+qa-r=0$$$$b^3-pb^2+qb-r=0$$$$c^3-pc^2+qc-r=0$$

Add these equations to obtain:$$(a^3+b^3+c^3)-p(a^2+b^2+c^2)+q(a+b+c)-3r$$And substituting and rearranging we get $$(a^3+b^3+c^3)-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$

To follow the comment about solutions to the cubic, examine the equation $$x^3+3px+q=0$$ and set $x=u+v$ then $$(u+v)^3=3uv(u+v)+u^3+v^3$$ //Note that this is $(u+v)^3-u^3-v^3=3(-u)(-v)(u+v)$ – which is another form of the earlier identity used by lab b in his solution//

and this is equivalent to $$x^3-3uvx-u^3-v^3$$

If we set $p=-uv$ and $q=-u^3-v^3$, we have $u^3, v^3$ as roots of $$y^2+qy-p^3=0$$Using the quadratic formula and $x=u+v$ we recover the form of the roots which is originally given.

Here $p=1, q=-18$ and the roots of the quadratic are $$\frac {18\pm\sqrt{324+-4}}{2}=9\pm \sqrt{80}$$