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? 

Suppose $\sqrt 3 + \sqrt[3]2$ were rational; that is,
$\sqrt 3 + \sqrt[3]2 = r \in \Bbb Q: \tag 1$
then
$\sqrt[3]2 = r - \sqrt 3; \tag 2$
we cube:
$2 = r^3 - 3r^2\sqrt 3 + 3r(\sqrt 3)^2 - (\sqrt 3)^3, \tag 3$
or
$2 = r^3 - 3r^2 \sqrt 3 + 9r - 3\sqrt 3, \tag 4$
or
$2 = r^3 + 9r - (3r^2 + 3)\sqrt 3, \tag 5$
or
$\sqrt 3 = \dfrac{2 - r^3 -9r}{3r^2 + 3} \in \Bbb Q, \tag 6$
which contradicts the fact that
$\sqrt 3 \notin \Bbb Q; \tag 7$
therefore, (1) is false.  $OE\Delta$.
