Proving irrationality of $\sqrt[3]{3}+\sqrt[3]{9}$ I need to prove $$\sqrt[3]{3}+\sqrt[3]{9}$$
is irrational, I assumed
$$\sqrt[3]{3}+\sqrt[3]{9} = \frac{m}{n}$$
I cubed both sides and got
$$\sqrt[3]{3}+\sqrt[3]{9} = \frac{m^3-12n^2}{9n^3}$$
I tried setting $$\frac{m^3-12n^2}{9n^3} = \frac{m}{n}$$ but that led me nowhere. so what can I do?

Let $a=\sqrt[3]{3}$ and $b=\sqrt[3]{9}$, and suppose that $a+b=r$ is rational. Then $$r^3 =(a+b)^3 = a^3+3ab(a+b)+b^3 = 12+9(a+b)=12+9r$$
So $r$ is a solution of equation $x^3-9x-12=0$ which has no rational root.
