<s>[INST] Computing $\sqrt[4]{28+16 \sqrt 3}$ I want to compute following radical
$$\sqrt[4]{28+16 \sqrt 3}$$
For that, I first tried to rewrite this in terms of exponential. 
$$(28+16\cdot 3^{\frac{1}{2}})^{\frac{1}{4}}$$
We know that $ 28 = 2 \cdot 7^{\frac{1}{2}}$
$$(2 \cdot 7^{\frac{1}{2}}+16\cdot 3^{\frac{1}{2}})^{\frac{1}{4}}$$
However, I'm stuck at this step. Could you assist me?
Regards
 [/INST] To find the fourth root, we need to use Bill Dubuque's denesting algorithm for square roots twice.
First time round, we can find that $\sqrt{\text{norm}} = \sqrt{28^2 - 3 \cdot 16^2} = \sqrt{16} = 4$. Subtracting out the norm gives $24 + 16 \sqrt{3}$, and when you divide this by $\sqrt{\text{trace}} = \sqrt{2 \cdot 24} = 4 \sqrt{3}$, we get $\frac{8 \sqrt{3}\sqrt 3}{4 \sqrt 3} + \frac{16 \sqrt3}{4 \sqrt3} = 2\sqrt{3}+4 = 4 + 2\sqrt{3}$.
Second time round, we can find that $\sqrt{\text{norm}} = \sqrt{4^2 - 3 \cdot 2^2  -8} = \sqrt{4} = 2$. Subtracting out the norm gives $2 + 2\sqrt{3}$, and when you divide this by $\sqrt{\text{trace}} = \sqrt{4} = 2$, you get $1 + \sqrt{3}$.
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