Irreducibility of $X^6+X^3+1$ in $\mathbb{Q}[X]$

Question

Could anyone advise me on how to prove $X^6+X^3+1$ is irreducible in $\mathbb{Q}[X] \ ?$

I'm thinking of substituting $X=Y+1$ into the equation, do some tedious computations to simplify and use Eisenstein's criterion. May I know if that is the correct approach?

Thank you.

Answer 1

If you don't want to do the multiplication, write $X^9-1=(X^6+X^3+1)\cdot(X^3-1)$. Now use your $X=Y+1$ trick and reduce mod 3: $$\begin{align} (Y+1)^9-1&=\left((Y+1)^6+(Y+1)^3+1\right)\cdot((Y+1)^3-1)\\ Y^9+1^9-1&=\left((Y+1)^6+(Y+1)^3+1\right)\cdot(Y^3+1^3-1)\\ Y^9&=\left((Y+1)^6+(Y+1)^3+1\right)\cdot Y^3\\ Y^6&=(Y+1)^6+(Y+1)^3+1\\ \end{align}$$ which tells you that all terms besides $Y^6$ have coefficient divisible by 3 (note that the Frobenius automorphism was used for the second line). Now plug $X=1$ into your original to find that the constant term is exactly three and then Eisenstein is good to go. The upshot to this method is that it generalizes to when you need to show, say, that $X^{78}+X^{75}+\cdots + X^3+1$ is irreducible.

Answer 2

$$(x+1)^6+(x+1)^3+1=x^6+6x^5+15x^4+21x^3+18x^2+9x+3$$

Well, Eisenstein is now a piece of cake, uh?