Are cube roots evenly distributed modulo primes?

Question

Are the cube roots of two integers chosen from a uniform distribution between $1$ and $p-1$ inclusive, $p$ prime, essentially evenly distributed? Note that I will use a $p$ such that $p$ is not equivalent to $2$ modulo $3$.

In other words, I'm trying to ensure that if I pick a random number between $1$ and $p-1$ I will have an equal chance of that number being the cube of some integer.

I've searched the site for answers, but my search did not come up with anything, and I don't know enough theory, except for maybe trying to read more on cubic reciprocity.

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I've found a similar result for square roots here.

Answer 1

One of your question is when we fix two (distinct cube free) positive integers $a,b$ and we choose randomly uniformly a prime $p\le N$ are the probabilities that $a$ and $b$ are cubes $\bmod p$ roughly equal (as $N$ gets large).

The answer is yes, because Chebotarev theorem tells us the asymptotic $$\#\{ p \le N, x^3-a\bmod p \text{ has a root }\}\sim \frac{|H|}{|G|} \frac{N}{\log N}$$ where $G=Gal(\Bbb{Q}(a^{1/3},\zeta_3)/\Bbb{Q}), |G|=6$ and $H$ is the set of $\sigma \in G$ such that $\sigma(a^{1/3})=a^{1/3}$, ie. $|H| = 2$ independently of $a$.