<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] This is an elementary but very good question. We have $$k = 2\sqrt{3} + 3\sqrt[3]{2} - 1$$ so that $$\sqrt[3]{2} = \frac{k + 1 - 2\sqrt{3}}{3}$$ and if $k$ is rational then it means that that $\sqrt[3]{2}$ is a quadratic irrationality i.e. it is a root of a quadratic equation with rational coefficients.
Hardy proves in "A Course of Pure Mathematics" that $\sqrt[3]{2}$ is not the root of such a quadratic equation. I show his method here. Let us suppose on the contrary that $\sqrt[3]{2}$ is a root of the equation $$ax^{2} + bx + c = 0\tag{1}$$ where $a \neq 0 \neq c$ and $a, b, c$ are integers. Now $\sqrt[3]{2}$ is also a root of $$x^{3} - 2 = 0\tag{2}$$ Multiplying $(1)$ by $2$ we get $$2ax^{2} + 2bx + 2c = 0$$and using $(2)$ we get $$2ax^{2} + 2bx + cx^{3} = 0$$ or $$cx^{2} + 2ax + 2b = 0\tag{3}$$ Multiply $(1)$ by $c$ and $(3)$ by $a$ and subtract to get $$(bc - 2a^{2})x + c^{2} - 2ab = 0$$ Now $x = \sqrt[3]{2}$ is irrational and hence the above equation is possible only when $c^{2} - 2ab = 0$ and $bc - 2a^{2} = 0$. This means that $c^{2} = 2ab$ and $bc = 2a^{2}$ so that $b^{2}c^{2} = 4a^{4}$ or $2ab^{3} = 4a^{4}$ or $b^{3} = 2a^{3}$. This is not possible unless $a = 0, b = 0$ because $\sqrt[3]{2}$ is irrational. We have reached a contradiction and hence $k$ is irrational.
Note: In terms of abstract algebra the above is a very elementary proof that polynomial $x^{3} - 2$ is irreducible over $\mathbb{Q}$ and $[\mathbb{Q}(\sqrt[3]{2}): \mathbb{Q}] = 3$. A proof more in line with techniques of abstract algebra is done via the concept of GCD. The GCD of polynomials in $(1)$ and $(2)$ also has $\sqrt[3]{2}$ as its root and since it is irrational it follows that the GCD has degree $2$ so that the polynomial in $(1)$ is the GCD. Hence $ax^{2} + bx + c$ is a factor of $x^{3} - 2$ and thus $x^{3} - 2$ has a linear factor so that $(2)$ has a rational root and thus the contradiction.
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