What are the conditions on $a, b, c$ so that $x^3+ax^2+bx+c$ is bijective? I would like to find the conditions on $a$, $b$, $c$ so that function $$f(x)=x^3+ax^2+bx+c$$ is bijective. 
I thought about resolving the equation
$$x^3+ax^2+bx+c=y$$
but I didn't succeed. And our math teacher told us that we cannot prove that a function is bijective by proving that this function is strictly increasing or decreasing. 
Thanks for your help!
Marie 

Surjectivity is clear, because a third degree equation always has at least a real root.
Suppose $f(x)=f(y)$, with $x\ne y$; then
$$
(x^3-y^3)+a(x^2-y^2)+b(x-y)=0
$$
that becomes
$$
(x-y)(x^2+xy+y^2+a(x+y)+b)=0
$$
and so $x^2+xy+y^2+a(x+y)+b=0$.
Set $s=x+y$ and $p=xy$: then $s^2-4p>0$. We have $s^2+as+b-p=0$, so
$$
s^2-4s^2-4as-4b>0
$$
or
$$
3s^2+4as+4b<0
$$
This is only possible if the discriminant of this polynomial in $s$ is positive: indeed, if the discriminant is $\ge0$, the inequality $3s^2+4as+4b\ge0$ holds for every $s$, contrary to the assumption that for the particular $s=x+y$ the $<0$ inequality holds. Thus we obtain
$$
a^2-3b>0
$$
Can you finish up?

 Conversely, suppose $a^2-3b\le0$. Then, for every $s$, we have $3s^2+4as+4b\ge0$, so $s^2-4(s^2+as+b)\le0$. If $f(x)=f(y)$, and we set $s=x+y$, $p=xy$, we have either $x=y$ or $s^2-4p\le0$, but this implies  $x=y$ again.

