Easy way to determine the primes for which $3$ is a cube in $\mathbb{Q}_p$? 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.

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). 
