How does one simplify the expression $\sqrt[3]{2 \sqrt{2}}\left(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\right)$ I don't know how to solve this problem. What I know is $\sqrt[3]{2 \sqrt{2}}=\sqrt{2}$. But I don't know how to continue.

$\sqrt[3]{2\sqrt 2}=\sqrt 2$ is correct. Since we have $$2\pm \sqrt 3=\frac{4\pm 2\sqrt 3}{2}=\frac{(\sqrt 3\pm 1)^2}{2}\Rightarrow \sqrt{2\pm \sqrt 3}=\frac{\sqrt 3\pm 1}{\sqrt 2},$$
we have
$$\sqrt 2\left(\sqrt{2-\sqrt 3}+\sqrt{2+\sqrt 3}\right)=\sqrt 2\left(\frac{\sqrt 3 -1}{\sqrt 2}+\frac{\sqrt 3+1}{\sqrt 2}\right)=\sqrt 3-1+\sqrt 3+1=2\sqrt 3.$$
