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 $\sqrt[3]3+\sqrt[3]9=r$.
Thus, since for all reals $a$, $b$ and $c$ we have:
 $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc),$$ we obtain:
$$3+9-r^3+9r=0.$$
Now, let $r=\frac{m}{n},$ where $m$ and $n$ are naturals with $gcd=1.$
Thus, $$m^3-9n^2m-12n^3=0,$$ which says that $m$ is divisible by $3$.
Let $m=3m'$, where $m'$ is a natural number.
Thus, $$9m'^3-9n^2m'=4n^3,$$ which says that $n$ is divisible by $3$, which is a contradiction.
