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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!

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    $\begingroup$ This seems like a lot of work. We know $x$ satisfies a cubic polynomial over $\mathbb Q$,namely $x^3-3$. If it also satisfies a quadratic then the quadratic and the cubic would have to a common factor, which is clearly not the case (since the cubic has no rational roots). $\endgroup$ – lulu Sep 8 '18 at 11:56
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    $\begingroup$ Notice that after raising both sides to the power of 6, your next step is to take the square root of both sides. Why not just raise to the power of 3 to begin with? $\endgroup$ – Théophile Sep 8 '18 at 12:09
  • $\begingroup$ @Théophile You are right, no reason not to just raise to the power of 3. Is this proof correct? $\endgroup$ – Omer Sep 8 '18 at 12:11
  • $\begingroup$ The proof looks correct to me. $\endgroup$ – saulspatz Sep 8 '18 at 12:19
  • $\begingroup$ Yes, it looks correct; the reasoning stands. Now, as lulu said, it is a lot of work and can be greatly simplified. Even if you keep the general structure here, you can save some writing in several places: apart from the power of 6 that I mentioned, also look at $1+\frac{4m}n$. This is clearly rational, and it's a bit cumbersome to write out, so why not replace it with something else? Indeed, you call it $k$, but why not make that substitution earlier, on the previous line? But let's be consistent and call it something like $\frac{m'}{n'}$ instead of $k$ to show that it is rational. $\endgroup$ – Théophile Sep 8 '18 at 12:20
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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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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.

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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.)

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