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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 $x^4-6x+3$. $a$ 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?

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?