Let $$\alpha = \sqrt[3]{2+\frac{10}{3\sqrt{3}}}\quad\text{and}\quad\beta=\sqrt[3]{2-\frac{10}{3\sqrt{3}}}$$ We have $\alpha\beta =\frac{2}{3}$. Suppose $\alpha+\beta =r$. Then $$r^3 = \alpha^3+\beta^3 + 3\alpha\beta(\alpha+\beta)=4+3\times\frac{2}{3}r$$ $$\therefore\quad r^3-2r-4=0\implies (r-2)(r^2+2r+2)=0$$ The only real root to this equation is $2$. Hence $\alpha+\beta = r=\boxed{2}$. By Vieta's theorem, $\alpha$ and $\beta$ are the two roots of equation $$t^2 -2t +\frac{2}{3}=0$$ \indent Solving this equation and noticing $\alpha > \beta$ yields: $$\alpha = 1+\frac{\sqrt{3}}{3}\text{and}\quad\beta=1-\frac{\sqrt{3}}{3}$$