## Is this irreducible?

Hi,
I need to figure out whether or not the polynomial
$$x^6-2x^3-1$$
is irreducible (over Q).
I don't think Eisenstein works in this case, and performing modulo 2 on this i get $$x^6-1$$
which is reducible over F2.
Any ideas? Incidently, if i let y=x^3, then i get
$$y^2-2y -1$$
which is irreducible over Q.....but i'm not sure if that means anything about the original polynomial.
 PhysOrg.com science news on PhysOrg.com >> City-life changes blackbird personalities, study shows>> Origins of 'The Hoff' crab revealed (w/ Video)>> Older males make better fathers: Mature male beetles work harder, care less about female infidelity
 Sure enough, Eisenstein doesn't work as there is no prime $$p$$ such that $$p\mid -1$$. However if you reduce modulo 3, you get (i hope) $$x^6-[2]x^3-[1] = x^6+[-2]x^3+[-1] = x^6+x^3+[2]$$ and this polynomial in $$Z_2[x]$$ seems to be irreducible (at a first try I couldn't factor it in 2 polynomials of smaller degree, but you can try yourself). So if that polynomial is irreducible in $$Z_2[x]$$ then $$x^6-2x^3-1$$ is irreducible in $$Q[x]$$.
 Hmm..okay. I was hoping I wouldn't have to resort to brute force hehe OK, so I have shown that the polynomial $$x^6+x^3+2$$ has no linear or quadratic factors over F3, but how did you show it can't factor into cubic terms? I don't really know the irreducible cubic polynomials over F3, and it seems like there might be quite a few of them?

Recognitions:
Homework Help