Prove that $X^6 - 10X^3 +8$ is irreducible over $\mathbb{Q}$.

Question

Prove that $X^6 - 10X^3 +8$ is irreducible over $\mathbb{Q}$.

Answer 1

Hint: It is enough to show that your polynomial is the minimal polynomial of $\sqrt[3]{5+\sqrt{17}}$.

Obviously $p(x)=x^2-10x+8$ is the minimal polynomial of $\alpha=5+\sqrt{17}$, whose conjugate is $\overline{\alpha}=5-\sqrt{17}$, and we have $[\mathbb{Q}(\alpha):\mathbb{Q}]=2$. If we set $\beta=\alpha^{1/3}$, then $[\mathbb{Q}(\beta):\mathbb{Q}(\alpha)]$ is either $1$ or $3$, depending on $x^3-\alpha$ being irreducible or not over $\mathbb{Q}(\alpha)$. A third-degree polynomial is irreducible iff it has no linear factors, so the problem boils down to checking if $5+\sqrt{17}$ is a cube in $\mathbb{Q}(\sqrt{17})$. That cannot happen, for many reasons. For instance, in $\mathbb{F}_{13}$ we have that $17$ is a quadratic residue (since $17\equiv 4$) but $5\pm 2$ is not a cubic residue, since: $$(5\pm 2)^{\frac{13-1}{3}}\not\equiv 1\pmod{13}.$$ As an alternative, we may check that the factorizations of $p(x)=x^6-10x^3+8$ over $\mathbb{F}_3$ and $\mathbb{F}_{13}$ are not compatible, so the conclusion is the same as above: $p(x)$ is irreducible over $\mathbb{Q}$.