Is $x^6 + bx^3 + b^2$ irreducible?

Question

Let $b\in \mathbb{Q}^*$ be rational number. We factorise $x^9-b^3\in \mathbb{Q}[x]$ and obtain $$x^9-b^3=(x^3-b)(x^6+bx^3+b^2).$$

Is the polynomial $x^6+bx^3+b^2$ irreducible?

If $b=1$ we get a cyclotomic polynomial, so yes. The same holds when $b$ is a cube. Is it true in general? It seems true because the roots of the polynomials are the $9$-th cube roots of $b^3$ and it seems that we cannot "separate" them, but is it true?

Answer 1

The answer is given by Theorem $3.1$ here:

Theorem 3.1. Let $h(x)=x^2+bx+c\in \Bbb Z[x]$ be irreducible. Then $$ f(x):=h(x^3)=x^6+bx^3+c $$ is reducible if and only if $c=n^3$ and $b=m^3-3mn$ for some integers $m,n$.

Answer 2

It's easy to factor it over $\mathbb{C}$, and then see that no product of those factors is a polynomial over $\mathbb{Q}$. Let $\omega$ be a primitive ninth root of unity, and let $\beta$ be the unique real cube root of $b$. Then,

$$p(x) = x^6+bx^3+b^2 = (x-\beta \omega)(x-\beta \omega^2)(x-\beta \omega^4)(x-\beta \omega^5)(x-\beta \omega^7)(x-\beta \omega^8)$$

Now note that if $p$ is reducible, one of the factors has degree $\le 3$. Clearly the degree is not $1$. If it is $2$, then we have a product which is irreducible over $\mathbb{Q}$,

$$(x-\omega^n\beta)(x-\omega^m \beta) = x^2 - (\omega^n + \omega^m)\beta x + \omega^{m+n}\beta^2$$

but for the constant coefficient to even be real, we have to have $m$ congruent to $-n$ mod 9 - then $\beta(\omega^n + \omega^{-n}) = 2\beta \cos(2n\pi/9)$, which is not rational unless $\cos(2n\pi/9)$ is, i.e. unless $n$ is a multiple of 3.

So we are factoring into two degree 3 polynomials, but considering the constant coefficient of one of those polynomials, it is of the form $\omega^{m+n+k}b$: again, in order to be real, we have to have $m+n+k$ equal to 0 mod 9. the only way that can happen is if they are all 1 or all 2 mod 3. But then the $x^2$ coefficient has a factor of $\omega(1+\omega^3+\omega^6) = -\omega$ (or $-\omega^2$) which also isn't real.