From N to infinity

The way I did the problem was to compute the probability that the primitive 2nd, 3rd, 4th, and 6th roots of unity were roots of $P$ (other than $1,$ which cannot be a root of $P(x),$ these are the only roots of unity whose minimal polynomials have degree $\le 3$ ). It turned out that the only way a 6th root of unity could be a root of $P$ was if a 3rd or 2nd root already was, so I could ignore that. The conditions on the exponents for the other three were conditions mod 2,3,4 respectively, so the middle condition was independent of the first and last. The first and last conditions also turned out to be mutually exclusive.

In other words, if $p_i$ denoted the probability that a primitive $i$ th root of unity ( $i = 2,3,4$ ) was a root of $P,$ then the probability that at least one of them is a root is $P_2 + P_3 + P_4 - P_2P_3 - P_3P_4.$ It's easy to compute $P_2 = 5/16,$ $P_3 = 10/81,$ $P_4 = 25/256.$ (Ok, I cheated and did $P_4$ by a computer search of exponents mod $4.$ But the other two are easy to derive from first principles.)

Anyway, this gives an answer of $10015/20736 \approx 0.482976466.$ But I have no idea how to prove that if $Q(x)$ is a cubic polynomial with nonzero constant term and integer coefficients which is not divisible by any of $x+1, x^2+x+1, x^2+1, x^2-x+1,$ then the probability that $Q(x)$ divides $P(x)$ goes to $0$ as $n \to \infty.$ Does anyone have an explanation?