Let k is integer which is not a cube. If there exists a prime p [respectively infinitely many primes p] congruent to 3 modulo 8 and such that k is not a cube modulo p? Thanks in advance for proof or counterexample!
Your condition $p \equiv 3 \pmod 8$ is unusual, everything is a cubic residue $\pmod 3$ and $\pmod q$ where $q \equiv 2 \pmod 3.$ However, Timo seems to have this well in hand.
This is just a bit of culture. A rational number is a square if and only if it is a square in every $\mathbb Q_p$ for all $p \leq \infty.$
I got a bit frustrated thinking about the analogous statement for cubes, but it works. In GSS it is proved that, if a polynomial in one variable with integer coefficients has prime degree, if it is reducible in all $\mathbb Q_p,$ then it is reducible in $\mathbb Q.$ In particular, the polynomial $x^3 -k$ is irreducible in $\mathbb Q$ if $k$ is not a cube. As the degree is 3, a prime, $x^3 -k$ is irreducible in some $\mathbb Q_p,$ meaning that there is no linear factor and no root, as $3 = 1 + 2.$ Finally, by Hensel's lemma, $k$ is not a cubic residue $\pmod p.$
Language in the article (mentioning Chebotarev density) suggests that the direction I use, prime degree, was known for quite some time, that the news in the article is about composite degree.
The laughter may now commence.