Proving $\sqrt3 + \sqrt[3]{2}$ to be irrational In a test I tried to solve recently I came across the following question:

Prove $$\sqrt3 + \sqrt[3]{2}$$ is irrational

I tried proving it by saying it is equal to some rational number $$\sqrt3 + \sqrt[3]{2} = \frac{m}{n}$$ and reaching a contradiction. 
I tried squaring / cubing both sides and that didnt work. So how do I go about solving this problem? 

It's possible to find a polynomial that has $\sqrt{3}+\sqrt[3]{2}$ as a root, along with several other variations from choosing different square and cube roots.
$$\left((x-\sqrt{3})^3-2\right)\left((x+\sqrt{3})^3-2\right)=0$$
$$\left(x^3-3\sqrt{3}x^2+9x-3\sqrt{3}-2\right)\left(x^3+3\sqrt{3}x^2+9x+3\sqrt{3}-2\right)=0$$
$$x^6-9x^4-4x^3+27x^2-36x-23 = 0$$
By the rational root theorem, the only possible rational roots of that polynomial are $-23,-1,1,$ and $23$. $\sqrt{3}+\sqrt[3]{2}$ is between $1+1=2$ and $2+2=4$, so it isn't any of them. Also, as is easily verified, none of those possible roots actually are roots. The values at $\pm 1$ are $1-9-4+27-36-23=-44$ and $1-9+4+27+36-23=36$, while at $\pm 23$ the $x^6$ term is much larger than everything else combined.
This is not the easy way, of course.
