<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] We have
$\displaystyle 2 + \frac{10}{9} \sqrt{3} = 1 + \sqrt{3} + 1 + \frac{1}{9} \sqrt{3}$
$\displaystyle = 1 + \frac{3}{\sqrt{3}} + \frac{3}{\sqrt{3}^2} + \frac{1}{\sqrt{3}^3}= \bigl(1 + \frac{1}{\sqrt{3}}\bigr)^3$
Similarily, we have $2-\frac{10}{9} \sqrt{3} = \left(1-\frac{1}{\sqrt{3}}\right)^3$. Hence
$\displaystyle \sqrt[3]{2+\frac {10} 9\sqrt 3}+\sqrt[3]{2-\frac {10} 9\sqrt 3} ~ = ~ \bigl(1 + \frac{1}{\sqrt{3}}\bigr) +\bigl(1-\frac{1}{\sqrt{3}}\bigr) = 2$.
PS: I have to admit that the first line is a little bit unmotivated. You need some experience in order to see these transformations (whereas verification is trivial). Alternatively, you can hope that both summands of this complicated sum actually lie in $\mathbb{Q}(\sqrt{3})=\{p+q \sqrt{3} : p,q \in \mathbb{Q}\}$, make the Ansatz $2+\frac{10}{9} \sqrt{3} = (p+q \sqrt{3})^3=\dotsc$ and solve for $p,q$, which gives $p=1$ and $q=\frac{1}{3}$ as a possible solution. The answer by lab bhattacharjee explains how to calculate the sum more "mechanically". 
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