Reduce the radical:

Question

From: Lumbreras Editors

---

So I proceeded:

$ \left(\sqrt{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}+\sqrt[4]{6 }\right)^4=\left(\sqrt{\sqrt{\frac{(1 +\sqrt{3})^2}{2}}+\sqrt{\frac{(1 -\sqrt{3})^2}{2}}}+\sqrt[4]{6}\right)^4\\ =\left(\sqrt{\frac{1+\sqrt{3}}{\sqrt{2}}+\frac{1-\sqrt{3}}{\sqrt{2}}}+\sqrt[4]{6}\right)^4 \left(\sqrt{\frac{2}{\sqrt{2}}} +\sqrt[4]{6}\right)^4 =\left(\sqrt[4]{2}+\sqrt[4]{2}\sqrt[4]{3}\right)^4\\ =2\left(1+\sqrt[4]{3}\right)^4 $

What else can be done?

Answer 1

Let

$$x=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}$$

$$x^2=2+\sqrt{3}+2-\sqrt{3}+2\sqrt{2+\sqrt{3}}\times \sqrt{2-\sqrt{3}}=6$$

$$x=\sqrt6$$

$$ (\sqrt{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}+\sqrt[4]{6 })^4=(2\sqrt[4]{6 })^4=96$$

Answer 2

Found your mistake!

$\sqrt{\frac{(1 -\sqrt{3})^2}{2}}\ne \sqrt{\frac{1-\sqrt{3}}{\sqrt 2}}$

Because the quantity under the square root in the RHS is negative

If you simplify like this

$\sqrt{\frac{(1 -\sqrt{3})^2}{2}}= \frac{\sqrt{3}-1}{\sqrt 2}$

everything works