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?

$$S=\displaystyle\sqrt[3]{2+\frac {10} 9\sqrt 3}+\sqrt[3]{2-\frac {10} 9\sqrt 3}$$
So, $$S^3=2+\frac {10} 9\sqrt 3+2-\frac {10} 9\sqrt 3+3\cdot \sqrt[3]{2+\frac {10} 9\sqrt 3}\cdot\sqrt[3]{2-\frac {10} 9\sqrt 3}\cdot S$$
$$\implies S^3=4+2S$$ as $(2+\frac {10} 9\sqrt 3)(2-\frac {10} 9\sqrt 3)=4-\frac{100\cdot3}{81}=\frac8{27}=(\frac23)^3$
$$\iff S^3-2S-4=0$$ 
$$\implies S^3-2^3-2(S-2)=0\implies (S-2)(S^2+S+2)=0$$
Observe that $2$ is the sole positive real root of the last eqaution
