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 Taxel Mar 10 at 3:37

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – uniquesolution, Cesareo, Carl Mummert, B. Goddard, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.

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

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

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.