Simple solving Skanavi book exercise: $\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}$ [closed]

Simple way to solve this exercise

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

HINT:

$$x^3=9+\sqrt{80}+9-\sqrt{80}+3\sqrt[3]{(9+\sqrt{80})(9-\sqrt{80})}(x)=18+3x$$

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

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 $$\left(\frac{3\pm\sqrt5}{2}\right)^3=9\pm\sqrt{80}.$$

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

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

The product

\begin{equation*} \left( u+\sqrt{v}\right) \left( u-\sqrt{v}\right) =u^{2}-v=\sqrt[3]{9+\sqrt{ 80}}\sqrt[3]{9-\sqrt{80}}=1 \end{equation*}

implies that

\begin{equation*} v=u^{2}-1, \end{equation*}

and

\begin{equation*} 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} \end{equation*}

A solution is

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

Since

\begin{eqnarray*} 4u^{3}-3u-9 &=&4\left( u-\frac{3}{2}\right) \left( u^{2}+\frac{3}{2}u+\frac{3 }{2}\right),\\ ( 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) , \end{eqnarray*}

the single solution of $(3)$ is

\begin{equation*} u=\frac{3}{2}.\tag{4} \end{equation*}

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

\begin{equation*} 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} \end{equation*}

and

\begin{eqnarray*} 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} \end{eqnarray*}