As the question says, how can I prove that $\sqrt[3]{2}+\sqrt[3]{4}$ is irrational?
I have tried setting it to be equal to $a$, and $\sqrt[3]{4}$ equal to $a^2$, but I haven't gotten anywhere. The solution does not have to use the above. Any help is appreciated. (I know this is a duplicate, but I haven't seen any answers that I have gotten a full solution from)
Let $\alpha=\sqrt[3]2+\sqrt[3]4$.
Then $\alpha^3=2+4+3\sqrt[3]{32}+3\sqrt[3]{16}=6+6\sqrt[3]4+6\sqrt[3]2=6+6\alpha.$
Apply the rational root theorem to $x^3-6x-6$.
Since, $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$$ and $$\sum_{cyc}(a^2-ab)=\frac{1}{2}\sum_{cyc}(a-b)^2=0$$ for $a=b=c$ only and $$\sqrt[3]2\neq\sqrt[3]4,$$ we obtain: $$\sqrt[3]2+\sqrt[3]4-x=0$$ is equivalent to $$2+4-x^3+3\sqrt[3]2\cdot\sqrt[3]4\cdot x=0$$ or $$x^3-6x-6=0.$$ But the polynomial $x^3-6x-6$ is irreducible by Eisenstein (just take $p=2$, for example): https://en.wikipedia.org/wiki/Eisenstein%27s_criterion
Thus, the polynomial $x^3-6x-6$ has no rational roots.
