Prove $x^6-6x^4+12x^2-11$ is irreducible over $\mathbb{Q}$ Extracted from Pinter's Abstract Algebra, Chapter 27, Exercise B1:

Let $p(x) = x^6-6x^4+12x^2-11$, which we can transform into a polynomial in $\Bbb{Z}_3[x]$:
  \begin{align*}
x^6+1
\end{align*}
  Since none of the three elements $0,1,2$ in $\Bbb{Z}_3$ is a root of the polynomial, the polynomial has no factor of degree 1 in $\Bbb{Z}_3[x]$. So the only possible factorings into non constant polynomials are
  \begin{align*}
x^6+1 &= (x^3+ax^2+bx+c)(x^3+dx^2+ex+f)
\end{align*}
  or
  \begin{align*}
x^6+1 &= (x^4+ax^3+bx^2+cx+d)(x^2+ex+f)
\end{align*}
  From the first equation, since corresponding coefficients are equal, we have
  \begin{align}
x^0:\qquad & cf &= 1 \tag{1} \\
x^1:\qquad & bf + ce &= 0 \tag{2} \\
x^2:\qquad & af + be + cd &= 0 \tag{3} \\
x^3:\qquad & c + f + bd + ae &= 0 \tag{4} \\
x^5:\qquad & a + d &= 0 \tag{5} \\
\end{align}
  From (1), $c = f = \pm1$, and from (5), $a + d = 0$. Consequently, $af + cd = c(a + d) = 0$, and by (3), $eb = 0$. But from (2) (since $c = f$), $b + e = 0$, and therefore $b = e = 0$. It follows from (4) that $c + f = 0$, which is impossible since $c = f = \pm1$. We have just shown that $x^6 + 1$ cannot be factored into two polynomials each of degree 3. 

For the second equation, however, $x^6+1=(x^2+1)^3$ in $\Bbb{Z}_3[x]$.  So we cannot say $p(x)$ is irreducible over $\Bbb{Q}$ because $x^6+1$ is irreducible over $\Bbb{Z}_3$.  What am I missing here?

Let $p(x) = x^6 - 6x^4 + 12x^2  - 11$
Substitute $x^2 = y$
$h(y) = y^3 - 6y^2 + 12y - 11$
Letting $g(y) = h(y+2)$
$$g(y) = (y+2)^3 -6(y+2)^2 + 12(y+2) - 11$$
$$g(y) = y^3 - 3$$
$g(y)$ is obviously irreducible, thus so is $h(y)$ and $p(x)$.
