How to prove $\sqrt3 + \sqrt[3]{2}$ is a irrational number? This is an exercise from R. Courant's book: How to prove $\sqrt3 + \sqrt[3]{2}$ is a irrational number?
The solution is to construct a equation to prove, but is there any other method to prove this, like by contradiction?
 A: If you're familiar with field extensions, you could do it like this:


*

*The extension $\mathbb{Q}(\sqrt{3})/\mathbb{Q}$ has degree two (minimal polynomial $X^2-2$)

*The extension $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ has degree three (minimal polynomial $X^3-2$)


This implies (by considering field towers) that the extension $\mathbb{Q}(\sqrt{3},\sqrt[3]{2})/\mathbb{Q}$ has degree 6, with basis $\{1,\sqrt[3]{2},(\sqrt[3]{2})^2,\sqrt{3},\sqrt{3}.\sqrt[3]{2},\sqrt{3}.(\sqrt[3]{2})^2\}$. So in particular, $\sqrt{3}+\sqrt[3]{2}$ is written out as an element of this basis and is not in $\mathbb{Q}$.
This might not make any sense to you (maybe one day it will), but you asked for another method, and this is a very general one.
A: Note that the intersection $ K = \mathbb{Q}(\sqrt[3]{2}) \cap \mathbb{Q}(\sqrt{3}) $ is a subfield of both fields, so it has degree 1 over $ \mathbb{Q} $ (since $ 2, 3 $ are coprime) and $ K = \mathbb{Q} $. If $ \sqrt{3} + \sqrt[3]{2} \in \mathbb{Q} $, then we would have $ \sqrt{3} \in K = \mathbb{Q} $, impossible.
Alternatively, consider $ \mathbb{Q}(\sqrt{3}) $ and note that the $\mathbb{Q}$-automorphism $ \sigma : \sqrt{3} \to -\sqrt{3} $ extends to an automorphism $ \varphi $ of the splitting field of $ X^3 - 2 $ over $ \mathbb{Q}(\sqrt{3}) $. It is then easily seen that $ \sqrt{3} + \sqrt[3]{2} = q $ cannot be rational, as $ \varphi $ can only map $ \sqrt[3]{2} $ to itself (otherwise we have that $\zeta_3$ is real) which implies that $ -\sqrt{3} + \sqrt[3]{2} = q $ as well, or $ \sqrt{3} = 0 $, impossible.
A: Assume that it is rational. Hence, $r=\sqrt{3} + \sqrt[3]{2}$ where $r = \frac{p}{q}$ for integral $p,\ q$ such that $q\not = 0$ and $p, \ q$ are coprime.
Hence, $$(r-\sqrt3)^3=2$$
$$\implies r^3-3\sqrt3-3\sqrt3r(r-\sqrt3) = 2 \\ \implies r^3-3\sqrt{3} -3\sqrt3r^2+9r=2 \\ \implies (r^3+9r) +\sqrt3(-3-3r^2)=2 $$
Since $r$ is rational, $r^3+9r$ has to be rational and $\sqrt3(-3-3r^2)$ has to be irrational. Since $2$ is purely rational, $$\sqrt3(-3-3r^2)=0 \\ \implies 3r^2=-3  $$
Hence, we arrive at a  contradiction. Thus, $r$ can not be rational. Hence, it is irrational.
A: Developing the hint by Winther. Assume it is rational $r=\sqrt{3}+\sqrt[3]{2}$. So we have
$$2=(r-\sqrt{3})^3=r^3-3r^2\sqrt{3}+9r-3\sqrt{3}$$
Regrouping and dividing by $3r^2+3\neq 0$ we get
$$\sqrt{3}={r^3+9r-2\over 3r^2+3}$$
This means $\sqrt{3}$ is rational. Contradiction
A: I do not know whether you know any Galois theory, but if you do: 
consider the splitting field over $\mathbb{Q}$ of $(X^2-3)(X^3-2)$, call this field $\Omega$. $\Omega/\mathbb{Q}$ is Galois and if $\zeta$ is a primitive root of unity of order 3, then there is a non-identity element $ \sigma \in \text{Gal}(\Omega/\mathbb{Q})$ that does not fix $\sqrt{3}+\sqrt[3]{2}$, namely take $\sigma$ the isomorphism $\Omega=\mathbb{Q}(\sqrt[3]{2}, \zeta\sqrt[3]{2}, \zeta^2 \sqrt[3]{2})(\sqrt{3}) \cong \mathbb{Q}(\sqrt[3]{2}, \zeta\sqrt[3]{2}, \zeta^2\sqrt[3]{2})(-\sqrt{3})= \Omega$ that fixes $\mathbb{Q}(\sqrt[3]{2}, \zeta\sqrt[3]{2}, \zeta^2 \sqrt[3]{2})$ and maps $\sqrt{3}$ to $-\sqrt{3}$. By the fundamental  theorem of Galois theory, $\alpha \in \Omega$ is in $\mathbb{Q}$ if and only if it is fixed by all elements of $\text{Gal}(\Omega/\mathbb{Q})$, but $\sigma(\sqrt{3}+\sqrt[3]{2}) = -\sqrt{3} + \sqrt[3]{2}$. It remains to show of course, that $\sqrt{3} \notin \mathbb{Q}(\sqrt[3]{2}, \zeta \sqrt[3]{2}, \zeta^2 \sqrt[3]{2})$. This follows from the fact that if it did, $S_3$ would have a normal subgroup of order 2, because $\text{Gal}(\mathbb{Q}(\sqrt[3]{2}, \zeta \sqrt[3]{2}, \zeta^2 \sqrt[3]{2})/\mathbb{Q}) \cong S_3$ and $\mathbb{Q}(\sqrt{3})/\mathbb{Q}$ is Galois, which is not the case.
If you do not know any Galois theory, I would like you to know that Galois theory makes a lot of these kind of questions easier to answer, and I would encourage you to learn it! 
