<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] In general, there's not much you can do with an expression like that. But if you know that the answer is in the form $a+b\sqrt 3$ for integers $a$ and $b$, then you can solve it like this:
Step 1 
($a+b\sqrt 3)^4=28+16\sqrt 3$. So suppose ($a+b\sqrt 3)^2=c+d\sqrt 3$. Then $(c+d\sqrt 3)^2=28+16\sqrt 3$, i.e.  
$$(c^2+3d^2)+2cd\sqrt 3=28+16\sqrt 3$$
$c^2+3d^2$ and $2cd$ are integers, so this can only be true if $c^2+3d^2=28$ and $2cd=16$.
Then $d=8/c$, so $c^2+\dfrac{192}{c^2}=28$. Multiplying by $c^2$ and rearranging, we get a quadratic in $c^2$:
$$c^4-28c^2+192=0$$
This factors as $(c^2-12)(c^2-16)=0$. Since $c$ is an integer, we must have $c=\pm4$, and so $d=\pm2$. So we have two possible solutions: $\pm(4+2\sqrt 3)$. (You should try squaring this to check that you get the expected result $28+16\sqrt 3$). Pick the positive solution $4+2\sqrt 3$ (if you choose the negative solution, you will find that Step 2 fails).
Step 2 
Now you need to solve $(a+b\sqrt 3)^2=4+2\sqrt 3$. You should be able to do this yourself now, using exactly the same method as Step 1.
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