TITLE: Easy way to determine the primes for which $3$ is a cube in $\mathbb{Q}_p$?
QUESTION [9 upvotes]: This is a qual problem from Princeton's website and I'm wondering if there's an easy way to solve it:
For which $p$ is $3$ a cube root in $\mathbb{Q}_p$?
The case $p=3$ for which $X^3-3$ is not separable modulo $p$ can easily be ruled out by checking that $3$ is not a cube modulo $9$. Is there an approach to this that does not use cubic reciprocity? If not, then I'd appreciate it if someone would show how it's done using cubic reciprocity. I haven't seen good concrete examples of it anywhere.
EDIT: I should have been more explicit here. What I really meant to ask was how would one find all the primes $p\neq 3$ s.t. $x^3\equiv 3\,(\textrm{mod }p)$ has a solution? I know how to work with the quadratic case using quadratic reciprocity, but I'm not sure what should be done in the cubic case.

REPLY [6 votes]: As noted in the comments, the question comes down to:

For which primes $p\gt 3$ is $3$ a cubic residue modulo $p$?

This is answered in detail in Franz Lemmermeyer's Reciprocity Laws, Chapter 7 ("Cubic Reciprocity"). 
If $p\equiv 2\pmod{3}$, then the order of the units modulo $p$ is prime to $3$, so every element is a cube; thus, $3$ is a cube modulo $p$ for all primes $p\equiv 2\pmod{3}$. 
If $p\equiv 1\pmod{3}$, then one can write $4p = L^2 + 27M^2$ for integers $L$ and $M$, and $3$ is a cubic residue modulo $p$ if and only if $M\equiv 0\pmod{3}$ (Proposition 7.2 in Lemmermeyer).