$x^6+3ax^4+3x^3+3ax^2+1$ is irreducible in $\Bbb Q$, where $a$ is a positive integer. $x^6+3ax^4+3x^3+3ax^2+1$ is irreducible in $\Bbb Q$, where $a$ is a positive integer.
How to prove it? Clealy, $f$ has no rational roots, so how to derive contradiction if $f(x)=g(x)h(x)$, where the degree of $g,h$ are $2,4$ (or $3,3$) respectively?
 A: Write $f(x)$ for $x^6 + 3ax^4 + 3x^3 + 3ax^2 + 1$ and suppose that we may write $f(x) = g(x) h(x)$ with $g, h \in \Bbb Q[x]$ both monic with degree $> 1$.
Since all roots of $$ are algebraic integers, it is clear that $g$ and $h$ both have integer coefficients.
We reduce the equality mod $3$. This gives $\bar f = \bar g \bar h$, where $\bar f = x^6 + 1\in \Bbb F_3[x]$.
We may factorize $\bar f$ over $\Bbb F_3$ as $\bar f = (x^2 + 1)^3$. Therefore, it is only possible that $g$ has degree $2$ and $h$ has degree $4$ (or vice versa). Moreover, we know that $g(x) \equiv x^2 + 1\pmod 3$.
Hence $g(x)$ is of the form $x^2 + 3bx + 1$ for some integer $b$. It follows that $h(x)$ is of the form $x^4 - 3bx^3 + cx^2 - 3bx + 1$ for some integer $c$, as $f$ has coefficient $0$ in degree $1$ and $5$.
Comparing coefficients, we get:
\begin{eqnarray}
c - 9b^2 + 1 &=& 3a\\
b(c - 2) &=& 1
\end{eqnarray}
The second equation only leads to two possibilities: $(b, c) = (1, 3)(-1,1)$. Plugging in them into the first equation, we find that in both cases the value of $a$ is not an integer. This finishes the proof.
