<s>[INST] 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?
 [/INST] 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 $x^3-3$.
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