Proving polynomial over Q is irreducible 
Prove that if $a$ and $b$ are odd then the polynomial $$x^3+ax+b$$is irreducible over $\mathbb{Q}$

I would be very much thankful if someone could help me with this one.
 A: A more direct approach might be to show that it is irreducible modulo $p$. Once we've shown that, it is surely irreducible over $\mathbb Z$ and hence $\mathbb Q$ as well (because the leading coefficient is $1$; by the rational root theorem, any rational root is integral). Since we are given $a,b$ are odd, the obvious choice is to take modulo $2$, where we have, in $\mathbb Z_2$,
$$x^3+ax+b=x^3+x+1.$$
Suppose this is reducible, then it has a root over $\mathbb Z_2$, and this must be either $0$ or $1$. But $0^3+0+1\neq0$ and $1^3+1+1=1\neq0$, so neither of the only two elements in $\mathbb Z_2$ are a root. We conclude that it is irreducible over $\mathbb Z_2$, hence over $\mathbb Q$ as well.
A: Say it is reducibile, then it has to have rational root and it has to be even an integer because leading coeficient is $1$. So we have $$(x^2+cx+d)(x+r) = x^3+ax+b$$
where $c,d,r\in \mathbb{Z}$. Since $rd = b$ and thus $r,d$ are both odd.
Now we have also $c+r =0$ (so $c$ is odd) and $cr+d =a$. A contradiction.
