The real number $b:=\frac{\sqrt{2a}+\sqrt{4\sqrt{a^2-3}-2a}}{2}$, where
\begin{equation*} a:=\sqrt[3]{\frac{9+\sqrt{65}}{4}}+\frac{1}{\sqrt[3]{\frac{9+\sqrt{65}}{4}}}=\frac{\sqrt[3]{18+2\cdot\sqrt{65}}}{2}+\frac{2}{\sqrt[3]{18+2\cdot\sqrt{65}}}, \end{equation*}
is a solution of equation $x^4-6x+3=0$. Eisenstein's criterion yields that $p(x)=x^4-6x+3$ is irreducible over $\mathbb{Q}$, which means that $b$ is algebraic of degree 4. Solving $x^4-6x+3=0$ leads to the cubic equation $y^3-3y-\frac{9}{2}$ of which $a$ is a solution. Therefore , $a$ cannot be constructed with ruler and compass.
How do you show (without Galois theory) that $b$ cannot be constructed with ruler and compass?