<s>[INST] Simple solving Skanavi book exercise: $\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}$ Simple way to solve this exercise
$$
x = \sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}
$$
 [/INST] 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*}
</s>