<s>[INST] Is this proof correct (Rationality of a number)? Is $\sqrt[3] {3}+\sqrt[3]{9} $ a rational number? My answer is no, and there is my proof. I would like to know if this is correct: 
Suppose this is rational. So there are positive integers $m,n$ such that $$\sqrt[3]{3}+\sqrt[3]{9}=\sqrt[3]{3}(1+\sqrt[3]{3})=\frac{m}{n}$$ 
Let $x=\sqrt[3]{3}$. We get $x^2+x-\frac{m}{n}=0 \rightarrow x=\frac{-1+\sqrt{1+\frac{4m}{n}}}{2}$.

 We know that $x$ is irrational and that implies $\sqrt{1+\frac{4m}{n}}$ is irrational as well (Otherwise $x$ is rational). Write $x=\sqrt[3]{3}$, multiply both sides by $2$ and then raise both sides to the power of $6$ to get: 
$$24^2=\left(\left(\sqrt{1+\frac{4m}{n}}-1\right)^3\right)^2\rightarrow 24=\left(\sqrt{1+\frac{4m}{n}}-1\right)^3=\left(1+\frac{4m}{n}\right)\cdot \sqrt{1+\frac{4m}{n}}-3\left(1+\frac{4m}{n}\right)+3\sqrt{1+\frac{4m}{n}}-1$$
Let $\sqrt{1+\frac{4m}{n}}=y,1+\frac{4m}{n}=k $. 
We get: $25=ky-3k+3y\rightarrow y=\frac{25+3k}{k+3}$, So $y$ is rational. But we know $y$ is irrational (again, otherwise $\sqrt[3]{3}$ is rational)  which leads to a contradiction. So the answer is No, $\sqrt[3]{3}+\sqrt[3]{9}$ is irrational. 
Is this proof correct? Is there another way to prove this? Thanks!
 [/INST] Alternatively, denote $x=\sqrt[3] {3}+\sqrt[3]{9}$ and cube it to get a cubic equation with integer coefficients:
$$x^3=3+3^2(\sqrt[3]3+\sqrt[3]{9})+9 \Rightarrow \\
x^3-9x-12=0.$$
According to the rational root theorem, the possible rational roots are: $\pm (1,2,3,4,6,12)$. However, none of them satisfies the equation. Hence, $x$ is irrational. 
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