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?

Let $x=\sqrt[3]{2+\frac {10}{9}\sqrt 3}, y=\sqrt[3]{2-\frac {10}{9}\sqrt 3}$. Then 
$$
x^3+y^3=4,\quad xy=\frac{2}{3}.
$$
The rest is yours.
