The number $b:=\frac{\sqrt{2a}+\sqrt{4\sqrt{a^2-3}-2a}}{2}$ with 
 $a:=\frac{\sqrt[3]{18+2\cdot\sqrt{65}}}{2}+\frac{2}{\sqrt[3]{18+2\cdot\sqrt{65}}}$ is a root of the irreducible polynomial $x^4-6x+3$.The algebraic number $a$ is a root of the irreducible polynomial $2x^3-6x-9$ and therefore not constructible with ruler and compass. 


I claim that $b$ isn't constructible either. If this is the case, how can this be shown?



Example of an algebraic number of degree 4 that is not constructible 
As explained in the comments, one needs to check if the splitting field of $x^4-6x+3$ has degree equal to a power of $2$ or not. The resolvent cubic of this polynomial equals $x^3-12x+36$ which is irreducible (over $\mathbb{Q}$), because it has no root modulo $7$. Hence the splitting field in question contains a subfield of degree $3$, so it cannot have degree equal to a power of $2$. This shows that $b$ is not constructible. You can read about the resolvent cubic [here](http://math.berkeley.edu/~serganov/114/cubqart.pdf).


