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.

For odd primes $q \equiv 2 \pmod 3,$ the cubing map is a bijection, 3 is always a cube $\pmod q.$
For odd primes $p \equiv 1 \pmod 3,$ by cubic reciprocity, 3 is a cube   $\pmod p$ if and only if there is an integer representation
$$  p = x^2 + x y + 61 y^2,  $$ or $4p=u^2 + 243 v^2.$ In this form this is Exercise 4.15(d) on page 91 of Cox. Also Exercise 23 on page 135 of Ireland and Rosen. The result is due to Jacobi (1827).
For more information when cubic reciprocity is not quite good enough, see Representation of primes by the principal form of discriminant $-D$ when the classnumber $h(-D)$ is 3 by Richard H. Hudson and Kenneth S. Williams, Acta Arithmetica (1991) volume 57  pages 131-153.
