Prove that $\sqrt[3]{2} + \sqrt[3]{4}$ is irrational.
My steps so far: I found that the polynomial $y^3-6y-6=0$ has roots $\sqrt[3]{2} + \sqrt[3]{4}.$ Can I use this to prove that $\sqrt[3]{2} + \sqrt[3]{4}$ is irrational? If so, how? I was thinking of using Proof by Contradiction, but I'm not so sure.
Actually, $\sqrt[3]2+\sqrt[3]4$ is not a root of that polynomial. But it is a root of $x^3-6 x-6$. By the rational roots theorem, the only rational roots that that polynomial can have are $\pm1$, $\pm2$, $\pm3$, and $\pm6$. Since none of them is actually a root, your number is irrational.