Step 1

Let \(\displaystyle{p}{\left({x}\right)}={x}^{{{6}}}-{6}{x}^{{{4}}}+{12}{x}^{{{2}}}-{11}\), 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

\(\displaystyle{x}^{{{6}}}+{1}={\left({x}^{{{3}}}+{a}{x}^{{{2}}}+{b}{x}+{c}\right)}{\left({x}^{{{3}}}+{\left.{d}{x}\right.}^{{{2}}}+{e}{x}+{f}\right)}\)

or

\(\displaystyle{x}^{{{6}}}+{1}={\left({x}^{{{4}}}+{a}{x}^{{{3}}}+{b}{x}^{{{2}}}+{c}{x}+{d}\right)}{\left({x}^{{{2}}}+{e}{x}+{f}\right)}\)

From the first equation, since corresponding coefficients are equal, we have

1) \(\displaystyle{x}^{{{0}}}:\quad{c}{f}={1}\)

2) \(\displaystyle{x}^{{{1}}}:\quad{b}{f}+{c}{e}={0}\)

3) \(\displaystyle{x}^{{{2}}}:\quad{a}{f}+{b}{e}+{c}{d}={0}\)

4) \(\displaystyle{x}^{{{3}}}:\quad{c}+{f}+{b}{d}+{a}{e}={0}\)

5) \(\displaystyle{x}^{{{5}}}:\quad{a}+{d}={0}\)

From (1), \(\displaystyle{x}={f}=\pm{1}\), and from (5), \(\displaystyle{a}+{d}={0}\).

Consequently, \(\displaystyle{a}{f}+{c}{d}={c}{\left({a}+{d}\right)}={0}\), and by (3), \(\displaystyle{e}{b}={0}\). 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, \(\displaystyle{x}^{{{6}}}+{1}={\left({x}^{{{2}}}+{1}\right)}^{{{3}}}\) 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 \(\displaystyle{\mathbb{{{Z}}}}_{{{3}}}\).