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?
Best Answer
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.