Consider a prime $p$ such that the polynomial $X^3-2$ splits into linear factors over $\mathbb{F}_p$: $X^3-2 = (X-\alpha_p)(X-\beta_p)(X-\gamma_p)$. It seems reasonable to expect that (identifying $\alpha_p,\beta_p,\gamma_p$ with numbers in $\{0,1,\dots,p-1\}$) the ratios $\alpha_p/p,\beta_p/p,\gamma_p/p \in [0,1)$ should be uniformly distributed (this expectation is, at the very least, strongly hinted at in Equidistribution of Roots of a Quadratic Congruence to Prime Moduli by W. Duke, J. B. Friedlander and H. Iwaniec). To be a little more precise, I would expect that if we look at the set of points
$$S_N = \{ \alpha_p/p : p < N \} \cup \{ \beta_p/p : p < N \} \cup \{ \gamma_p/p : p < N \}$$ then for each interval $[x,y) \subset [0,1)$ the ratio $\#( S_N \cap [x,y))/\# S_N$ should converge to $y-x$ as $N \to \infty$. This, however, seems to be an open problem: the analogous question for quadratic polynomials was solved by Duke, Friedlander and Iwaniec, and as far as I'm aware the same problem for higher degrees is considered difficult.
I am curious if a weaker statement can be proved. Specifically, can we disprove that for each $\varepsilon > 0$ and all sufficiently large primes $p$ we have $\alpha_p/p \in [0,\varepsilon)$? In other words, can we show that there are infinitely many primes $p$ such that $X^3-2$ does not have a root modulo $p$ in $\{0,1,2,\dots, \lfloor\varepsilon p\rfloor\}$? By pigeonhole argument (using the fact that $n^3-2$ can be divisible by at most two primes larger than $n$) one can disprove that for almost all primes $p$ we have $\alpha_p/p \in [0, c/\log p)$ for a constant $c > 0$.
