Question is:
Let $f$ be an irreducible (in $Z[x]$) monic polynomial with integer coefficients and of odd degree greater than $1$. Suppose that the modules of the roots of $f$ are greater than $1$ and that $f(0)$ is a square-free number. Prove that the polynomial $g(x) = f(x^3)$ is also irreducible.
Here is my idea to approach:
Suppose contradict, $g(x)$ is reducible then $g(x)$ is a multiple of an irreducbile $h(x)$. It means $f(x^3)$ is a multiple of $h(x)$. Let $\omega=\dfrac{-1+i\sqrt{3}}{2}$ then $f(x^3)$ is a multiple of $h(\omega x)$ and $h(\omega^2 x)$. I'm stuck here. I wonder but can't proof the $h(x),h(\omega x) $ and $h(\omega^2x)$ are pairwise coprime polynomials.
Somebody can help me?? Thanks!