Is this proof correct (Rationality of a number)?

Question

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!

Answer 1

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.

Answer 2

Your proof is correct.

Let $\alpha = \sqrt[3]{3} + \sqrt[3]{9} = \sqrt[3]{3}(1+\sqrt[3]{3})$. We have

$$\alpha^3 = 3(1+\sqrt[3]{3})^3 = 3(3 + 3\sqrt[3]{3} + 3\sqrt[3]{9} + 3) = 12 + 9(\sqrt[3]{3} + \sqrt[3]{9}) = 12 + 9\alpha$$

Hence $\alpha^3 - 9\alpha -12 = 0$.

However, the polynomial $x^3-9x-12$ is irreducible over $\mathbb{Q}$ by the Eisenstein criterion for $p = 3$ so it cannot have any rational roots. Therefore $\alpha$ is not rational.

Answer 3

As everyone else has said, your proof is correct but more laborious than necessary. Here's an approach that I think may minimize the amount of calculation.

Let $x=\root3\of3$, so we're interested in $y=x^2+x$. The fact that $x$ is a cube root suggests thinking about things involving $x^3$, and if you're lucky the following identity, involving both $x^3$ and $y$, may come to mind: $x^3-1=(x-1)(x^2+x+1)$. Aha! The left-hand side of this is a rational number, namely 2. And if $y$ is rational then the second factor on the right is also rational. But that would require $x-1$, and hence $x$, to be rational, which it certainly isn't. So $y$ can't be rational after all, and we're done.

(An easy generalization of the argument shows that things like $\root5\of7+\root5\of{7^2}+\root5\of{7^3}+\root5\of{7^4}$ are always irrational.)