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?

Showing that $\displaystyle\sqrt[3]{2+\frac {10} 9\sqrt 3}+\sqrt[3]{2-\frac {10} 9\sqrt 3} = 2$ is the same as showing that $2\pm\frac {10} 9\sqrt 3$ has a cube root of the form $1 \pm a\sqrt 3$ (with rational $a$).
$(1\pm a\sqrt 3)^3 = (1+9a^2)\pm(3a+3a^3)\sqrt 3$, and you quickly check that with $a = 1/3$ you simultaneously get $1+9a^2 = 2$ and $3a(1+a^2) = \frac{10}9$, hence $2 \pm \frac{10}9\sqrt 3 = (1\pm\frac 13\sqrt 3)^3$, and your original expression simplifies to $1+\frac 13\sqrt 3+1-\frac 13\sqrt 3=2$
