How does one simplify the expression $\sqrt[3]{2 \sqrt{2}}\left(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\right)$

Question

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.

Answer 1

Hint : $$ \left(\sqrt{a}\pm\sqrt{b}\ \right)^2=a+b\pm2\sqrt{ab} $$ or $$ \frac{\sqrt{a}\pm\sqrt{b}}{\sqrt2}=\sqrt{\frac{a+b}2\pm\sqrt{ab}}. $$ For example, we have $\sqrt{2-\sqrt{3}}$, then $\dfrac{a+b}2=2$ and $ab=3$.

Answer 2

$\sqrt[3]{2\sqrt 2}=\sqrt 2$ 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 $$\sqrt 2\left(\sqrt{2-\sqrt 3}+\sqrt{2+\sqrt 3}\right)=\sqrt 2\left(\frac{\sqrt 3 -1}{\sqrt 2}+\frac{\sqrt 3+1}{\sqrt 2}\right)=\sqrt 3-1+\sqrt 3+1=2\sqrt 3.$$

Answer 3

$$\sqrt[3]{2 \sqrt{2}}=\sqrt2$$ $$\sqrt{2-\sqrt{3}}=\frac{\sqrt{3}-1}{\sqrt 2},\sqrt{2+\sqrt{3}}=\frac{\sqrt{3}+1}{\sqrt 2}$$

$$\sqrt[3]{2 \sqrt{2}}\left(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\right)=\sqrt[3]{2^{3/2}}\left(\frac{\sqrt{3}-1}{\sqrt 2}+\frac{\sqrt{3}+1}{\sqrt 2}\right)=$$ $$=\sqrt 2\frac{\sqrt3-1+\sqrt3+1}{\sqrt 2}=2\sqrt3$$