How many prime $p$ that not satisfy with any integer $x$ in $x^3-2028x+2018\equiv 0 \pmod{p^3}$ 
From the topic , how many prime $p$ that cannot find any integer $x$ to satisfy
  $$x^3-2028x+2018\equiv 0 \pmod{p^3}$$

I try  to start with $x^3-2028x+2018\equiv 0 \ \ \left (mod \ p   \right )$ and  put some small $p$ like $2,3,5,7$  and then notice which $x$ can be applied ,  it's tedious task for me to do with more bigger $p$.
But I don't know how to deal with $p^3$  and  I'm not sure that there are infinite $p$ to match at least one $x$ or not?
i'm appreciate that you could hint me for some theory to start with. Thank you to any reply. 
 A: Very generally, if $f$ is a nonlinear irreducible polynomial over $\mathbb{Z}$, then there are always infinitely many prime numbers $p$ such that $f$ has no roots mod $p$.  By the Frobenius density theorem (which you can find an excellent brief overview of here), it suffices to show that some element of the Galois group $G$ of $f$ has no fixed points (when acting on the roots of $f$).  But the set of elements of $G$ that have a fixed points is just the union of the conjugates of the subgroup $H$ that fixes one particular root, since $G$ acts transitively on the roots.  Since a finite group cannot be covered by the conjugates of a proper subgroup, some element of $G$ has no fixed points.
In particular, then, your cubic $f(x)=x^3-2028x+2018$ is irreducible over $\mathbb{Z}$ since it is monic and has no integer roots, so there are infinitely primes $p$ such that it has no roots mod $p$, and hence also no roots mod $p^3$.
There might be a more elementary way to prove this for this particular polynomial, but if it's just some random polynomial you chose, I wouldn't expect there to be one.
