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?
 A: 
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?

You are entirely correct here. What you are missing is that the preceding argument shows that $p$ is not a product of two cubics in $\Bbb{Q}[x]$. After all, if it were, then in such a factorization
$$p=(x^3+ax^2+bx+c)(x^3+dx^2+ex+f),$$
all coefficients are integers by Gauss' lemma, and hence this reduces to a factorization into cubics in $\Bbb{F}_3[x]$. But you have just shown that no such factorization exists.
So it remains to show that $p$ is not the product of a quadratic and a quartic in $\Bbb{Q}[x]$, and there are infinitely many other primes to try. For example, in $\Bbb{F}_7[x]$ you have
$$x^6-6x^4+12x^2-11=x^6+x^4+5x^2+3.$$
A quick check shows that this polynomial has no roots in $\Bbb{F}_7$. Now proceed as before, expanding
$$x^6+x^4+5x^2+3=(x^4+ax^3+bx^2+cx+d)(x^2+ex+f),$$
to show that no such factorization exists in $\Bbb{F}_7[x]$.

If you are comfortable with a little more abstract algebra, here is an approach that does not require such ad hoc calculations. First it is easy to see that in $\Bbb{F}_3[x]$ the polynomial $p$  factors as
$$p=x^6+1=(x^2+1)^3,$$
where $x^2+1\in\Bbb{F}_3[x]$ is irreducible. It follows that every irreducible factor of $p$ in $\Bbb{Q}[x]$ has even degree. Now note that $p=h(x^2)$ where $h:=x^3-6x^2+12x-11\in\Bbb{Q}[x]$. A quick check shows that $h$ has no roots in $\Bbb{F}_7$, and therefore it is irreducible in $\Bbb{F}_7[x]$. This means the subring of the quotient ring $\Bbb{F}_7[x]/(p)$ generated by $x^2$ is a cubic field extension of $\Bbb{F}_7$, and therefore $p$ has an irreducible cubic or sextic factor. In the latter case $p$ is irreducible in $\Bbb{F}_7[x]$, and hence in $\Bbb{Q}[x]$ and we are done.
If $p$ has an irreducible cubic factor in $\Bbb{F}_7[x]$, then this is the reduction of an irreducible factor of $p$ in $\Bbb{Q}[x]$. As we saw before, the degree of this factor is even, so it is either quartic or sextic. Again, if it is sextic then $p$ is irreducible in $\Bbb{Q}[x]$ and we are done. If it is quartic then its reduction in $\Bbb{F}_7[x]$ is the product of a cubic and a linear factor. But $p$ has no roots in $\Bbb{F}_7$ because $p=h(x^2)$ and $h$ has no roots in $\Bbb{F}_7$, a contradiction.
A: Update: The answer is wrong but see my comment!
I think the reasoning should be like the following. As $p(x)$ has integer coefficients and is monic every zero of $p$ that lies in $\mathbb{Q}$ is also integer. But every integer zero of $p$ must divide the absolute term which is 11. Therefore, it could only be $\pm1$ or $\pm11$. Neither is a solution. The explanation above shows that $p$ cant be the product of two cubic polynomials. So if $p$ were reducible it had an irreducible quadratic monic polynomial with as factor which would have itself two zeros of the form $\pm\sqrt{q}+r$ whose square is in $\mathbb{Q}$. But then $p$ has only even powers of $x$ so substituting $x^2\rightarrow{}y$ gives a cubic polynomial which is either irreducible or has a linear factor. But then the same reasoning as above applies and since neither $\pm1$ nor $\pm11$ are zeros of that polynomial we are done.
A: 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 $p(x)$.
