<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] Assuming $$\frac{p}{q}=2\sqrt{3}+3\sqrt[3]{2}$$
You will have :
$$54 = (\frac{p}{q} - 2\sqrt{3})^3$$
$$= \frac{p^3}{q^3}-6\frac{p^2}{q^2}\sqrt{3}+18\frac{p}{q}-24\sqrt{3}$$
So finally :
$$54 -\frac{p^3}{q^3}-18\frac{p}{q}= -6\frac{p^2}{q^2}\sqrt{3}-24\sqrt{3} = -6 (\frac{p^2}{q^2}+4)\sqrt{3}$$
which is impossible because $\sqrt{3}$ is irrational and $54 -\frac{p^3}{q^3}-18\frac{p}{q}$ and $-6 (\frac{p^2}{q^2}+4)$ are both rational
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