Prove x^{6}-6x^{4}+12x^{2}-11 is irreducible over \mathbb{Q}

Step 1
Let ${\displaystyle p\left(x\right)=x^6-6x^4+12x^2-11}$, whith we can transform into a polynomial in ${\displaystyle {\mathbb{Z}}_3\left[x\right]}$:
${\displaystyle x^6+1}$
Since none of the three elements 0,1,2 in ${\displaystyle {\mathbb{Z}}_3}$ is a root of the polynomial, the polynomial has no factor of degree 1 in ${\displaystyle {\mathbb{Z}}_3\left[x\right]}$. So the only possible factorings into non constant polynomials are
${\displaystyle x^6+1=(x^3+a{x}^2+bx+c)(x^3+{dx}^2+ex+f)}$
or
${\displaystyle x^6+1=(x^4+a{x}^3+b{x}^2+cx+d)(x^2+ex+f)}$
From the first equation, since corresponding coefficients are equal, we have
1) ${\displaystyle x^0:cf=1}$
2) ${\displaystyle x^1:bf+ce=0}$
3) ${\displaystyle x^2:af+be+cd=0}$
4) ${\displaystyle x^3:c+f+bd+ae=0}$
5) ${\displaystyle x^5:a+d=0}$
From (1), ${\displaystyle x=f=\pm 1}$, and from (5), ${\displaystyle a+d=0}$.
Consequently, ${\displaystyle af+cd=c(a+d)=0}$, and by (3), ${\displaystyle eb=0}$. But from (2) (since ${\displaystyle c=f}$), ${\displaystyle b+e=0}$, and therefore ${\displaystyle b=e=0}$. It follows from (4) that ${\displaystyle c+f=0}$ which is impossible since ${\displaystyle c=f=\pm 1}$.
We have just shown that ${\displaystyle x^6+1}$ cannot be factored into two polynomials each of degree 3.
For the second equation, however, ${\displaystyle x^6+1={(x^2+1)}^3}$ in ${\mathbb{Z}}_3[x]$. So we cnnot say ${\displaystyle p\left(x\right)}$ is irreducible over ${\displaystyle \mathbb{Q}}$ because ${\displaystyle x^6+1}$ is irreducible over ${\displaystyle {\mathbb{Z}}_3}$.

Let ${\displaystyle p\left(x\right)=x^6-6x^4+12x^2-11}$
Substitute ${\displaystyle x^2=y}$
${\displaystyle h\left(y\right)=y^3-6y^2+12y-11}$
Letteing ${\displaystyle g\left(y\right)=h(y+2)}$
$g(y)=(y+2{)}^3-6(y+2{)}^2+12(y+2)-11$
${\displaystyle g\left(y\right)=y^3-3}$
${\displaystyle g\left(y\right)}$ is obviously irreducible, thus so is ${\displaystyle h\left(y\right)}$ and ${\displaystyle p\left(x\right)}$.

Step 1
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 ${\mathbb{F}}_3[x]$ the polynomial p factors as
$p=x^6+1=(x^2+1{)}^3$
where $x^2+1\in {\mathbb{F}}_3[x]$ is irreducible. It follows that every irreducible factor of p in $\mathbb{Q}[x]$ has even degree. Now note that $p=h(x^2)$ where $h:=x^3-6x^2+12x-11\in \mathbb{Q}[x]$
A quick check shows that h has no roots in ${\mathbb{F}}_7$, and therefore it is irreducible in ${\mathbb{F}}_7[x]$. This means the subring of the quotient ring $\frac{\mathbb{F}[x]}{(p)}$ generated by $x^2$ is a cubic field extension of ${\mathbb{F}}_7$, and therefore p has an irreducible cubic or sextic factor. In the latter case p is irreducible in ${\mathbb{F}}_7[x]$, and hence in $\mathbb{Q}[x]$ and we are done.
If p has an irreducible cubic factor in ${\mathbb{F}}_7[x]$, then this is the reduction of an irreducible factor of p in $\mathbb{Q}[x]$.
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 $\mathbb{Q}[x]$ and we are done. If it is quartic then its reduction in ${\mathbb{F}}_7[x]$ is the product of a cubic and a linear factor. But p has no roots in ${\mathbb{F}}_7$ because $p=h(x^2)$ and h has no roots in ${\mathbb{F}}_7$, a contradiction.