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Prove that $\sqrt[3]{2} + \sqrt[3]{4}$ is irrational.

My steps so far: I found that the polynomial $y^3-6y-6=0$ has roots $\sqrt[3]{2} + \sqrt[3]{4}.$ Can I use this to prove that $\sqrt[3]{2} + \sqrt[3]{4}$ is irrational? If so, how? I was thinking of using Proof by Contradiction, but I'm not so sure.

In-finite
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  • The given expression can be written as $\sqrt[3] 2(\sqrt[3] 2+1) $. Knowing that $\sqrt[3] 2$ is irrational, the product of two irrational numbers is irrational, too. – SarGe Jun 14 '20 at 08:49
  • @Doubtnut "the product of two irrational numbers is irrational, too" Not so, e.g. $\sqrt{2}\cdot\sqrt{2}=2$. – J.G. Jun 14 '20 at 08:50
  • @Doubtnut No you are not right, for example $\pi$ and $e$ are both irrationals but it is still an open problem that $\pi\cdot e$ is whether irrational or not. – mertunsal Jun 14 '20 at 08:51
  • Maybe it might be said that (it is a rough conjecture feel free to refute me as you want) product of algebraic irrational numbers with different degrees are irrational. – mertunsal Jun 14 '20 at 08:53
  • In this case, $\sqrt[3]2$ and $\sqrt[3]2+1$ both have the same degree. @mertunsal22 – Angina Seng Jun 14 '20 at 09:13
  • @Angina Seng Yes, I didn't bind them with if and only if. The conjecture itself is nothing to do with the question of OP's but rather than the argument of Doubtnut. – mertunsal Jun 14 '20 at 09:15

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Actually, $\sqrt[3]2+\sqrt[3]4$ is not a root of that polynomial. But it is a root of $x^3-6 x-6$. By the rational roots theorem, the only rational roots that that polynomial can have are $\pm1$, $\pm2$, $\pm3$, and $\pm6$. Since none of them is actually a root, your number is irrational.

José Carlos Santos
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