Prove that $\sqrt 2 + \sqrt[3]2$ is not rational

Question

How do I prove that the following is not rational?

$$x=\sqrt 2 + \sqrt[3]2$$

To prove a simpler case like $\sqrt{2}=a/b$, I can raise both sides to the power of 2 and get $a^2=2b^2$, therefore both $a$ and $b$ must be even numbers which can't be true.

Answer 1

Let $$x=\sqrt 2 + \sqrt[3]2$$ then $$x-\sqrt 2 = \sqrt[3]2$$ $$\left(x-\sqrt 2 \right)^3=2$$ $$x^3-3x^2\left( \sqrt 2 \right)+3x(2)-2\left( \sqrt 2 \right)=2$$ $$\sqrt 2= \frac {x^3+6x-2}{3x^2+2}$$

$x$ cannot be rational because $\frac {x^3+6x-2}{3x^2+2}$ will then be rational and yet $\sqrt 2$ is irrational.