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It seems that it is irreducible over $\mathbb Q$, I tried to apply Eisenstein's criterion to $f(x+a)$ for some $a$, but it didn't work.

Also, I found a root $\alpha = \sqrt[3]{\frac{\sqrt{17}-3}{2}}$ and tried to prove that the degree of $\alpha$ over $\mathbb Q$ is 6, but I stuck in computation.

Is there any approach to prove or disprove the irreducibility of the above polynomial?

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    Set $y=x^2$ and use the rational root theorem on the new polynomial – 2017-01-21
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    That doesn't always work, @YoTengoUnLCD $x^4+4$ is reducible, but $x^2+4$ is irreducible. – 2017-01-21
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    Set $u=x^3$ and you get the quadratic $u^2+3u-2$ with roots $u=\dfrac{-3\pm\sqrt{17}}{2}=x^3$. So you will get two irrational real roots and four complex roots. – 2017-01-21
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    @JohnWaylandBales Think OP wants a proof that u^{1/3} is irrational – 2017-01-21
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    Although all roots of the polynomial are not rational, it doesn't guarantee the irreducibility of the polynomial. $x^4 +4 $ is an example. – 2017-01-21
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    If it was reducible, then it would be reducible to monic polynomials in $\mathbb{Z}$ by Gauss lemma. Then, the constant term of the factors must be $\pm 1, \pm 2$. Since you know all the roots, I think you can argue by the sizes of absolute values of the roots. – 2017-01-21
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    @JWL I just did it! Thank you so much! – 2017-01-21
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    Can anyone answer to the continuing question: http://math.stackexchange.com/q/2107565/380038 – 2017-01-21

4 Answers 4

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Its not hard to show that $[\mathbb{Q}(\alpha):\mathbb{Q}]=6$. We have $$[\mathbb{Q}(\alpha):\mathbb{Q}]=[\mathbb{Q}(\alpha):\mathbb{Q}(\sqrt{17})][\mathbb{Q}(\sqrt{17}):\mathbb{Q}]$$ and $$[\mathbb{Q}(\sqrt{17}):\mathbb{Q}]=2$$ so we must show that $$[\mathbb{Q}(\alpha):\mathbb{Q}(\sqrt{17})]=3.$$

This amounts to showing that $x^3=\frac{-3+\sqrt{17}}{2}$ has no solution in $\mathbb{Q}(\sqrt{17})$ Since $\sqrt{17}\mapsto -\sqrt{17}$ is an automorphism of this field we see that if there is an $x\in \mathbb{Q}(\sqrt{17})$ such that $$x^3=\frac{-3+\sqrt{17}}{2}$$ then there is a $y\in \mathbb{Q}(\sqrt{17})$ such that $$y^3=\frac{-3-\sqrt{17}}{2}$$ multiplying these two equations we get that $$(xy)^3=-2$$ And we know that $\mathbb{Q}(\sqrt[3]{-2}):\mathbb{Q}]=3$ a contradiction so there is no such $x$ and $[\mathbb{Q}(\alpha):\mathbb{Q}(\sqrt{17})]=3.$

Or alternatively you could take the norm to get $(N(x))^3=-2$ an equation in $\mathbb{Q}$ again a contradiction.

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Let $f = x^6 - 3x^3 - 2$.

Suppose $f$ is reducible over $\mathbb{Q}$.

Of the irreducible factors of $f$, let $g$ be one of least degree.

It follows that $\text{deg}(g) \le 3$.

By the rational root test, $f$ has no rational roots, hence $\text{deg}(g) > 1$.

Let $w$ be a root of $g$, and let $v = w^3$.

Then $v$ satisfies $v^2 - 3v - 2 = 0$, so $v = \frac{3}{2} \pm \frac{1}{2} \sqrt{17}$.

Since $[\mathbb{Q}(v):\mathbb{Q}]=2$ and $v \in \mathbb{Q}[w]$, it follows that $[\mathbb{Q}(w):\mathbb{Q}]$ is even, hence $\text{deg}(g) = 2$.

But then $[\mathbb{Q}(w):\mathbb{Q}(v)]=1$, so $w \in \mathbb{Q}(v)$.

Since, $v = w^3$, the norm of $v$ in $\mathbb{Q}(v)$ must be a perfect cube in $\mathbb{Q}$.

But the norm of $v$ in $\mathbb{Q}(v)$ is $\left(\frac{3}{2}\right)^2-17\left(\frac{1}{2}\right)^2 = -2$, which is not the cube of a rational number, so we have a contradiction.

It follows that $f$ is irreducible over $\mathbb{Q}$.

2

Note that

$$x^6+3x^3-2\equiv \begin{cases} (x^2+1)^3\mod 3\\ (x^3-4)(x^3-6)\mod13 \end{cases} $$

Since $-1$ is not a square mod $3$, the quadratic $x^2+1$ is irreducible mod $3$, and since neither $4$ nor $6$ is a cube mod $13$ (the cubes mod $13$ are $\pm1$ and $\pm5$), the cubics $x^3-4$ and $x^3-6$ are irreducible mod $13$. Therefore $x^6+3x^3-2$ is irreducible over $\mathbb{Q}$: any (monic) factor of degree $d$ would have to reduce to a factor of the same degree both mod $3$ and mod $13$.

Remark: Why look mod $13$? you may well ask. It's because $13$ has two good properties. First, the discriminant of the quadratic in $x^3$, $3^2+4\cdot2=17$, is a square mod $13$, which guarantees a factorization of the form $(x^3-a)(x^3-b)$. And second, $3\mid\phi(13)$, so not every number is a cube mod $13$. Beyond that, it was just dumb luck that $4$ and $6$ are both among the non-cubes.

1

A brute force method that will always work for arbitrary 6th degree polynomials in $\mathbb{Z}[x]$ (and factor it if it is reducible), involves reducing the polynomial modulo $x^2 - p x - q$, setting the result to zero, and then eliminating $p$ to obtain an equation for $q$ to check for quadratic factors. To check for cubic factors one reduces the polynomial modulo $x^3 - p x^2 - q x - r$, sets the result to zero and then eliminates $p$ and $q$ to obtain an equation for $r$. For the quadratic case, one obtains the equation:

$$q^{15} + 7 q^{12} - 62 q^9 + 124 q^6 - 56 q^3 -32 = 0 $$

Then $q$ must be an integer that divides $2$ but none of the possibilities satisfy the above equation, so there are no quadratic factors.

For the cubic case, one obtains the equation:

$$ \begin{split} &r^{20}+3 r^{19}-2 r^{18}-18 r^{17}-54 r^{16}+36 r^{15}+84 r^{14}+252 r^{13}-168 r^{12}+72 r^{11}\\ &+216 r^{10} -144 r^9-672 r^8-2016 r^7+1344 r^6-1152 r^5-3456 r^4+2304 r^3\\ &-512 r^2-1536 r+1024 = 0 \end{split} $$

No integer $r$ that divides $2$ satisfies this equation, so there aren't any cubic factors either therefore the polynomial is irreducible.