<s>[INST] Show that $\sqrt[3]{2+\frac {10} 9\sqrt 3}+\sqrt[3]{2-\frac {10} 9\sqrt 3}=2$ 
Find $\displaystyle\sqrt[3]{2+\frac {10} 9\sqrt 3}+\sqrt[3]{2-\frac {10} 9\sqrt 3}$.

I found that, by calculator, it is actually $\bf{2}$.
Methods to denest something like $\sqrt{a+b\sqrt c}$ seems to be useless here, what should I do?
 [/INST] Find $p$ and $q$ such that
\begin{cases}
-\frac{q}{2}=2\\[3ex]
\frac{p^3}{27}+\frac{q^2}{4}=\frac{100}{27}
\end{cases}
This of course gives $q=-4$ and
$$
\frac{p^3}{27}=\frac{100}{27}-4=-\frac{8}{27},
$$
so $p^3=-8$ and $p=-2$.
Now find the unique real root of the polynomial $x^3+px+q=x^3-2x-4$, which is of course $2$. Indeed the polynomial factor as
$$
x^3-2x-4=(x-2)(x^2+2x+2)
$$
and the second degree factor has no real roots.
This example clearly shows why Cardan's formulas have a very limited usefulness: they present the roots in a form that makes them unintelligible.
Here's Cardan's formula for third degree equations with only one real root: if $x^3+px+q=0$ and the discriminant
$$\frac{p^3}{27}+\frac{q^2}{4}>0$$
then the only real solution for the equation is
$$
\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{p^3}{27}+\frac{q^2}{4}}}
+
\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{p^3}{27}+\frac{q^2}{4}}}
$$
</s>