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?
 A: Here are two other takes.
Take 1
Let $\alpha = \sqrt[3]{3}+\sqrt[3]{9}$. Then $\alpha^3 = 9 \alpha + 12$.
By the rational root theorem, if $\alpha$ is rational, then $\alpha$ is an integer.
Now $1 < \sqrt[3]{3} < 2 $ and $2 < \sqrt[3]{9} < 3 $, and so $3 < \alpha < 5$. 
Since $x=4$ is not a root of $x^3 = 9 x + 12$, $\alpha$ is not an integer and so $\alpha$ is irrational.
Take 2
We have $\alpha = \beta+\beta^2$, where $\beta=\sqrt[3]{3}$.
If $\alpha$ were rational, then $\beta$ would be a root of a quadratic polynomial with rational coefficients. However, the polynomial with rational coefficients with least degree having $\beta$ as a root is $x^3-3$.
A: 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.
A: 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.
