An irreducibility problem. Here is a problem from one of the Harvard quals question paper(spring 2011 I think) which I could not figure it out for a long time. 
Prove that for any positive integer $a$, the polynomial $f(x)=x^6+3ax^4+3x^3+3ax^2+1$ is irreducible.
I tried to use Eisenstein criterion but I couldn't find a suitable transformation to $f$ so that Eisenstein can be applied and I don't know any other criterion other than this( If we need some other criterion to solve this please mention it).
Please give your ideas/hints to solve this.
Thanks.
 A: Idea:
If $f(x)= P(x)Q(x)$, with $P,Q$ monic, then in $Z_3[x]$ you have
$P(x)Q(x)= x^6+1=(x^2+1)^3$. 
$x^2+1$ is irreducible over $Z_3[x]$, thus the only possibility (up to renaming) is $P(x) \equiv x^2+1 \mod 3$ and $Q(x) \equiv (x^2+1)^2 \mod 3$.
I think the problem should follow easily from here, express $P(x)=x^2+1+3P_1(x)$, $Q(X)=(X^2+1)^2+3Q_1(X)$, with $\deg(P_1) \leq 1, \deg(Q_1) \leq 3$  and plug it back into the first equation....
A: I think it should be sufficient to consider the reduction mod 2. If $a$ is even, then the polynomial reduces to $x^6+x^3+1$. If $a$ is odd, then the polynomial being considered reduces to $x^6+x^4+x^3+x^2+1$. Clearly, neither of these polynomials have constant factors. If they can be reduced then it must be as a product of an irreducible quadratic and an irreducible quartic, two irreducible cubics, or as a product of three irreducible quadratics. Notice that in $\mathbb{Z}/(2)[x]$, the only irreducible quadratic is $x^2+x+1$. But neither polynomial has $x^2+x+1$ as a possible factor. Thus, if the two polynomials are reducible modulo 2, then they must be a product of two irreducible cubics. There are only two irreducible cubics in $\mathbb{Z}/(2)[x]$: $x^3+x^2+1$ and $x^3+x+1$. A quick computation shows this to be impossible. 
Since the polynomial is irreducible modulo 2 for every $a$, the polynomial $x^6+3ax^4+3x^3+3ax^2+1$ is irreducible for every $a$.
A: Here is what I tried. This does not cover the case of quadratic-times-quartic. 
First, we can use synthetic division to determine that there are no linear factors. For example, $f(-1) = 6a -1 = 0 \Rightarrow a \notin \mathbb{Z}^+$ 
Then I considered $$f(x) = (x^3 + bx + c + 1)(x^3 -bx^2 - cx + 1)$$
The justification of these coefficients is the absense of the fifth- and first-degree terms in $f$. From here, we see (if my calculations are correct!) that $3a = -b^2$, impossible for our values of $a$. 
Surely there is a more theoretic/less brute force approach, but I thought this was worth mentioning. 
