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

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.

• 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 8:49
• @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 8:50
• @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 8:51
• 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 8:53
• In this case, $\sqrt[3]2$ and $\sqrt[3]2+1$ both have the same degree. @mertunsal22 – Angina Seng Jun 14 '20 at 9:13

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.
• @MathIsFun123 Use Factor Theorem and calculate $p(\sqrt[3]{2} + \sqrt[3]{4})$. – Landuros Jun 14 '20 at 8:52
• Did you read the edited version of my answer? I wrote there that it is a root of $x^3-6x-6$. – José Carlos Santos Jun 14 '20 at 8:56