So we have a polynomial in the form: $ax^3+bx^2+cx+d$, where $a,b,c,d\in\mathbb{Q}$, $a\neq 0$. And this is irreducible over $\mathbb{Q}[x]$, but is of course reducible over $\mathbb{C}[x]$.
We have the three roots: $\alpha, \beta, \gamma\in(\mathbb{C}\setminus\mathbb{Q})$. And we know that $\gamma=\alpha\cdot\beta=\alpha\beta$
So far, I have tried writing out the Vieta’s Formula: $$-\frac{b}{a}=\alpha+\beta+\gamma=\alpha+\beta+\alpha\beta \\ \frac{c}{a}=\alpha\beta+\alpha\gamma+\beta\gamma=\alpha\beta+\alpha^2\beta+\alpha\beta^2=\alpha\beta(1+\alpha+\beta) \\ -\frac{d}{a}=\alpha\beta\gamma=\alpha^2\beta^2=(\alpha\beta)^2$$
I tried getting a contradiction from these but to no avail. Is there such a polynomial?
If $\gamma=\alpha\beta$, then $-\frac da=\alpha\beta\gamma=\gamma^2$. This means $\gamma$ is a root of the irreducible quadratic polynomial $ax^2+d$. A root of a quadratic cannot also be a root of an irreducible cubic.