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 
 A: 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.

A: All such maps $f:\>{\mathbb R}\to{\mathbb R}$ are surjective by the intermediate value theorem.
On the other hand, one has $f(x)-f(y)=(x-y)g(x,y)$ with
$$\eqalign{ g(x,y)&=(x^2+xy+y^2)+a(x+y)+b\cr
&=\left(x+{a\over3}\right)^2+\left(x+{a\over3}\right)\left(y+{a\over3}\right)+\left(y+{a\over3}\right)^2+b-{a^2\over3}\ .\cr}$$
The quadratic form
$$q(u,v):=u^2+uv+v^2=\left(u+{v\over2}\right)^2+{3\over4}v^2$$
is positive definite. It follows that $g(x,y)$ assumes only positive values if $b-{a^2\over3}>0$, hence $f$ is injective in this case. 
If $b-{a^2\over3}=0$ then $g\left(-{a\over3},-{a\over3}\right)=0$, and $g(x,y)>0$ for all other pairs $(x,y)$, in particular when $x\ne y$. This allows to conclude that $f$ is injective in this case as well.
If $b-{a^2\over3}<0$ then the equation $g(x,y)=0$ defines an ellipse $E$ in the $(x,y)$-plane. There are points $(x,y)\in E$ with $x\ne y$. For such $x$, $y$ one has $f(x)=f(y)$, hence $f$ is not injective in this case.
To sum it all up: The map $f$ is bijective iff $a^2\leq3b$.
A: If $f'(x)>0$ for every $x$ or ($f'(x)<0$ for every $x$) it is 1-1 function and also onto function ,so it can be bijective
$$f'(x)=3x^2+2ax+b\\3x^2+2ax+b>0 \to \Delta \leq 0\\(2a)^2-4(3)(b)\leq0 \to a^2\leq 3b $$
this condition is an answer . note that $c$ is not important in this case
A: Disclaimer: I started from a simple geometrical idea, but, unfortunately, it turned out messier than expected. This answer, therefore, just serves the purpose of showing that this can be done. Other answers are by far more elegant and are recommended over this one.
Let $f(x) = x^3+ax^2 + bx + c$. Since $y\mapsto y + c$ is bijective, we can assume that $c = 0$.
As others already noted, surjectivity is automatic, so we only need to deal with injectivity.
I will appeal to geometry at this point: 

$f$ is not injective if and only if there is a line parallel to
  $x$-axis that crosses the graph of $f$ at more than one point.

This tells us that:

$f$ is not injective if and only if there exists $y\in\Bbb R$ such
  that $f(x) - y$ has at least two different real roots.

Since non-real roots of polynomials over $\Bbb R$ come in conjugate pairs,

$f$ is not injective if and only if there exists $y\in\Bbb R$ such
  that $f(x) - y = (x-A)(x-B)(x-C)$ where $A,B,C$ are real
  numbers, not all equal.

Expanding we get $$x^3 + ax^2 + bx - y = x^3 -(A+B+C)x +(AB+BC+AC)x - ABC$$ so the question boils down to finding real $A,B,C$ such that \begin{array}{c c}\begin{align}-(A+ B+C)&=a\\ AB+BC+AC &= b\end{align} & \tag{1} \end{array} and then we set $y = ABC$.
We can now conclude that 

$f$ is injective if and only if $(1)$ has no real solutions or all the real solutions satisfy $A=B=C$.

Solving system $(1)$ in terms of $A$, we get $$B = \frac 1 2 (-a - A \pm \sqrt{- 3 A^2 - 2 a A +a^2 - 4 b})\\ C = -a - A - B\tag{2}$$
We can easily see that $(1)$ has no real solutions if and only if $B$ and $C$ from $(2)$ are not real for any real $A$, and that happens if and only if $- 3 A^2 - 2 a A +a^2 - 4 b<0,\ \forall A\in \Bbb R$. 
From the discriminant of $- 3 A^2 - 2 a A +a^2 - 4 b$ as a polynomial in $A$, we finally get condition that $(1)$ has no real roots if and only if $a^2 < 3b$.
Assume now that there is a real solution to $(1)$ such that $A = B = C$. Substituting in $(1)$ we get necessary condition $A = B = C = -a/3$ and $a^2 = 3b$.
Now we need to show that $a^2 = 3b$ implies that all the real solutions to $(1)$ satisfy $A = B = C$. But this is almost immediate: we must have $$- 3 A^2 - 2 a A +a^2 - 4 b\geq 0$$ but the discriminant of $- 3 A^2 - 2 a A +a^2 - 4 b$ is $0$ (because $a^2 = 3b$), so we can conclude that $A = -a/3$ is the only real that satisfies this, and also we get that in that case $A = B = C$. Thus, $a^2 = 3b$ implies that $A=B=C=-a/3$ is the unique real solution to $(1)$.
Collecting the pieces, we get that $f$ is injective if and only if $a^2 \leq 3b$.
A: In the following we use simple quadratic/parabola/discriminant theory learned in high school to show that the OP's $f$ is injective if and only if $a^2 \le 3b$.
Consider the two statements
$\tag 1 u \ne v \land (u-v)\big(u^2+(v+a)u + v^2 + av +b\big)=0 $
$\tag 2 u^2+(v+a)u + v^2 + av +b=0 \land a^2 \gt 3b$
Lemma: $\text{(1)} \implies \text{(2)}$.
Proof
Assume $\text{(1)}$ is true:
Then $u^2+(v+a)u + v^2 + av +b=0$. 
The quadratic function $g(x) = x^2+(v+a)x + (v^2 + av +b)$ satisfies $g(u) = 0$ and therefore, using quadratic/parabola/discriminant theory,
$\tag 3 (v+a)^2 -4(1)(v^2 + av +b) \ge 0$
or equivalently
$\tag 4 -3v^2 -2av + a^2 -4b \ge 0$
The quadratic function $h(x) =  -3x^2 -2ax + a^2 -4b$ satisfies $h(v) = 0$ and therefore, using quadratic/parabola/discriminant theory,
$\tag 5 (-2a)^2 - 4 (-3) (a^2 -4b) \ge 0$
or equivalently
$\tag 6 a^2 \ge 3b$
We claim that $a^2 = 3b$ is impossible. If it was true, the graph of $g$ would intersect the $x\text{-axis}$ at exactly one point and we would find that (after employing our theory yet again)
$\quad [v = -\frac{a}{3}] \land [u = v]$
contradiction our assumption that $u \ne v$. So the inequality is strict,  $\quad \blacksquare$
Using the lemma it is an algebraic/logic exercise (recall that $s-t$ divides $s-t$ taken to the $n^\text{th}$ power) to show that if a cubic
$\quad f(x)=x^3+ax^2+bx+c$
is not injective then $a^2 \gt 3b$.
The converse is also true. If $a^2 \gt 3b$ using parabola theory we can choose a $v$ such that $h(v) \gt 0$. But then we certainly have two distinct numbers $w$ to choose from such that $g(w) = 0$. At least one of these is not equal to $v$, and set $u$ to that the number.
So $\text{(1)}$ holds for these numbers $u$ and $v$ and therefore $f$ is not injective.
