<s>[INST] Proving that $2\sqrt 3+3\sqrt[3] 2-1$ is irrational 
Prove that $2\sqrt 3+3\sqrt[3] 2-1$ is irrational

My attempt:
$$k=2\sqrt 3+3\sqrt[3] 2-1$$
Suppose $k\in \mathbb Q$, then $k-1\in \mathbb Q$.
$$2\sqrt 3+3\sqrt[3] 2=p/q$$
I'm stuck here and don't know how to procced. I tried to do this:
$$\sqrt 3=\frac{p/q-3\sqrt[3] 2}{2}$$
contradiction, but I'm not at all sure about that. How should I proceed?
 [/INST] \begin{align*}
2\sqrt 3+3\sqrt[3] 2=\frac{p}{q} &\Rightarrow 3\sqrt[3] 2=\frac{p}{q}-2\sqrt 3\\
 &\Rightarrow 54=\left(\frac{p}{q}-2\sqrt 3\right)^3\\
 &\Rightarrow 54=\frac{p^3}{q^3}-6\frac{p^2}{q^2}\sqrt{3}+36\frac{p}{q}-24\sqrt{3}\\
 &\Rightarrow \sqrt{3}=\frac{p^3+36pq^2-54q^3}{6q(p^2+4q^2)} \in \mathbb{Q}\\
\end{align*}
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