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As the question says, how can I prove that $\sqrt[3]{2}+\sqrt[3]{4}$ is irrational?

I have tried setting it to be equal to $a$, and $\sqrt[3]{4}$ equal to $a^2$, but I haven't gotten anywhere. The solution does not have to use the above. Any help is appreciated. (I know this is a duplicate, but I haven't seen any answers that I have gotten a full solution from)

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  • $\begingroup$ One application of Fermat's Last Theorem is to show that $\sqrt[3]{2}$ is irrational. :) $\endgroup$ – user844292 Nov 2 '20 at 4:55
  • $\begingroup$ That application is discussed here $\endgroup$ – J. W. Tanner Nov 2 '20 at 22:39
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    $\begingroup$ Not only is this a duplicate. It is a duplicate of a duplicate. Shame on the high rep users who don't search (took me less than 5 seconds on Approach0). $\endgroup$ – Jyrki Lahtonen Nov 11 '20 at 10:58
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Let $\alpha=\sqrt[3]2+\sqrt[3]4$.

Then $\alpha^3=2+4+3\sqrt[3]{32}+3\sqrt[3]{16}=6+6\sqrt[3]4+6\sqrt[3]2=6+6\alpha.$

Apply the rational root theorem to $x^3-6x-6$.

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Since, $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$$ and $$\sum_{cyc}(a^2-ab)=\frac{1}{2}\sum_{cyc}(a-b)^2=0$$ for $a=b=c$ only and $$\sqrt[3]2\neq\sqrt[3]4,$$ we obtain: $$\sqrt[3]2+\sqrt[3]4-x=0$$ is equivalent to $$2+4-x^3+3\sqrt[3]2\cdot\sqrt[3]4\cdot x=0$$ or $$x^3-6x-6=0.$$ But the polynomial $x^3-6x-6$ is irreducible by Eisenstein (just take $p=2$, for example): https://en.wikipedia.org/wiki/Eisenstein%27s_criterion

Thus, the polynomial $x^3-6x-6$ has no rational roots.

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