Prove that $\sqrt{2} +\sqrt[3]{2}$ is irrational.
My attempt: Suppose $\sqrt{2} +\sqrt[3]{2}$ is rational then $\exists$ $x\in \mathbb{Q}$ such that $$\sqrt{2} +\sqrt[3]{2}=x$$ Rewriting the above equation as $$x-\sqrt{2}=\sqrt[3]{2}$$ cubing the above equation gives us $$x^3-3x^2\sqrt{2}+6x-2\sqrt{2}=2$$ This implies that $$\sqrt{2}=\frac{x^3+6x-2}{3x^2+2}\in\mathbb{Q}$$ but this is absurd . Therefore ,$\sqrt{2} +\sqrt[3]{2}$ is irrational. Does this look good ? Can this be done any other way.
This question is from chapter 2 of Spivak's Calculus. In there he's given a hint which says "start by working with the 6th power " of this expression.. I don't see how that helps . Can you show me how it can be done using the hint the Author has provided ? Thank you
Best Answer
You've got the right idea, but there are two (small) problems:
the equation $$x^3-3x^2\sqrt{2}+6x+2\sqrt{2}=2$$ is wrong, the last sign on the left should be $-$, as in $(-1)^3=-1$.
you should explain why it is possible to divide by $3x^2-2$. In fact, once you fix the first point I mentioned, it's going to be $3x^2+2$ and it's going to be much easier.