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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?

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

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  • $\begingroup$ Are you sure that is irrational? $\endgroup$ – uniquesolution Mar 9 at 10:56
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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 $x=\sqrt[3]p+p\sqrt[3]{p^2}$.

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

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