# Prove that $\sqrt[3]p+\sqrt[3]{p^5}$ is irrational if $p$ is prime [closed]

Prove that $$\sqrt[3]p+\sqrt[3]{p^5}$$ is irrational when $$p$$ is a prime.

First I suppose $$x=\sqrt[3]p+\sqrt[3]{p^5}$$. Cubing gives $$x^3=p+p^5+p^2x$$ And then what properties of prime, and how to test its irrationallity?

## closed as off-topic by uniquesolution, Cesareo, Carl Mummert, B. Goddard, Parcly TaxelMar 10 at 3:37

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• Are you sure that is irrational? – uniquesolution Mar 9 at 10:56

## 1 Answer

By Eisenstein's criterion, $$X^3-p$$ is irreducible over $$\Bbb Q$$. Therefore $$1$$, $$\sqrt[3]p$$ and $$\sqrt[3]{p^2}$$ are linearly independent over $$\Bbb Q$$.

Now observe that $$x=\sqrt[3]p+p\sqrt[3]{p^2}$$.

• At this level one should really justify the inference "Therefore...." – Bill Dubuque Mar 9 at 21:21