$x^3-3n^2x+n^3$ is irreducible over $\mathbb{Q}[x], \forall n \in \mathbb{N}$.

Show that $$p(x)=x^3-3n^2x+n^3$$ is irreducible over $$\mathbb{Q}[x], \forall n \in \mathbb{N}$$.

$$\textbf{My observations:}$$

The possible rational roots is the divisors of $$n^3$$. However, $$p(n),p(n^2),(n^3) \neq 0$$ Then the possible rational roots of $$p$$ divide $$n$$ (It doesn't help a lot).

I've tried to use the Eisenstein's-Criterion but it doesn't work...because I don't know $$n$$.

The other way is to show that this polynomial is irreducible in $$\mathbb{Z}_p[x]$$, for some $$p$$ prime. Are there some property about a cubic of a number?

Can you help me with a hint about that?

• It suffices to prove this for $n=1$. Aug 12 '20 at 5:45
• Ok, for $n=1$ is easy to show that $p(x)$ is irreducible in $\mathbb{Z}_3[x]$ so in $\mathbb{Q}[x]$ it is. My questions is why is it suffices? Aug 12 '20 at 5:48
• It is not irreducible in $\mathbb{Z}_3[x]$ since it is equal $(x+1)^3$ there.
– Sil
Aug 12 '20 at 6:06
• you're right, my other attempt is: the possible rational roots of $x^3-3x+1$ is $\pm1$... it's clear that $p(\pm 1) \neq 0$... so it is irreducible. Aug 12 '20 at 6:09

I assume $$n \neq 0$$.
The polynomial is reducible iff it has a rational root. Suppose we have $$x^3 - 3n^2 x + n^3 = 0$$. Then $$\frac{x^3}{n^3} - 3 \frac{x}{n} + 1 = 0$$. Then the polynomial $$x^3 - 3x + 1$$ has a rational root. But by the rational root theorem, such a root would have to be $$\pm 1$$; clearly, neither is a root.
• How $x^3-3x+1$ is irreducible by Eisenstein? Aug 12 '20 at 6:40
• The finishing should be like the following $\rightarrow$ Since $x^3-3x+1$ is monic, it has rational root means that is an integer. But for all Integers $n\leq0$ $n^3-3n+1>0$, for all $n>1$ $n^3-3n+1>0$ and $n=1$ gives the value $-1$. So no integer roots. Hence it is irreducible over the rationals. Aug 12 '20 at 6:45