I am having trouble with this: it appears that $x^{12} - x^6 + 1$ has a root $\pmod p$ only when $p \equiv 1 \pmod{36},$ in which case it completely factors as twelve linear terms.
It is easy to show that any prime for which there is a root satisfies $p \equiv 1 \pmod{12},$ from $$ x^{12} - x^6 + 1 = (x^6-1)^2 + x^6 = (x^6)^2 - x^6 + 1 $$
There are plenty of interesting factorizations modulo other primes, it is just the all terms are degree two or higher.
Oh, this comes from an earlier question, here is the comment I made there: Find another sum of squares for $3^{12}-6^6+2^{12}$
Let me paste in some data; Meanwhile, the question is, does $x^{12} - x^6 + 1$ have a root $\pmod p$ only when $p \equiv 1 \pmod{36} \; \; ? \; \;$
jagy@phobeusjunior:~$ ./count_roots
prime
37 count 1 p mod 36 : 1
73 count 2 p mod 36 : 1
109 count 3 p mod 36 : 1
181 count 4 p mod 36 : 1
397 count 5 p mod 36 : 1
433 count 6 p mod 36 : 1
541 count 7 p mod 36 : 1
577 count 8 p mod 36 : 1
613 count 9 p mod 36 : 1
757 count 10 p mod 36 : 1
829 count 11 p mod 36 : 1
937 count 12 p mod 36 : 1
1009 count 13 p mod 36 : 1
1117 count 14 p mod 36 : 1
1153 count 15 p mod 36 : 1
The key here is to recognize the polynomial as a cyclotomic polynomial. From the sum/difference of cubes factorization, $(x^6+1)(x^{12}-x^{6}+1)=(x^{18}+1)$, and from the difference of squares factorization, $(x^{18}+1)(x^{18}-1)=x^{36}-1$. A root of this polynomial will have order dividing $36$. However, our original polynomial is actually the cyclotomic polynomial $\Phi_{36}(x)$, and so a root modulo $p$ will have order exactly $36$. The order of an element mod $p$ must divide $p-1$, and so $36\vert p-1.$