Show that the function $y \mapsto y^3 + ay$ is injective for $a < 0$? Show that the map $$y \mapsto y^3 + ay \ \ \ \ \ \ \  (y \in \mathbb R)$$ is injective where $a < 0$.
I see that the derivative is given by $\ \ 3y^2 + a$.
I've computed $3y^2 + a > 0 \iff y > \sqrt {\frac {|a|} 3}$ or $y < -\sqrt {\frac {|a|} 3}$. So the map is injective on the intervals $(-\infty, -\sqrt {\frac {|a|} 3})$, $(\sqrt {\frac {|a|} 3}, \infty)$ and I see that the image on these intervals are disjoint (right?).
How do I see that on the interval $[-\sqrt {\frac {|a|} 3},\sqrt {\frac {|a|} 3}]$, the map $y \mapsto y^3 + ay$ has image disjoint from the images of intervals above and that it is injective on this interval ? This would allow me to conclude that the map $y \mapsto y^3 + ay$ is injective ?
 A: I don't think it's true because for $a=-1$ you have the same value for $y=0$ and $y=1$.
A: You have the statement exactly backwards: The function $y \mapsto y^3 + ay$ is injective on all reals if and only if $a\ge 0$.
The derivative is indeed $3y^2+a$. If $a>0$ this is clearly always positive, so the function is strictly increasing and thus injective. If $a=0$ the function is $y^3$ which we know to be injective.
However, if $a<0$ the derivative at $y=0$ is negative and thus is decreasing in an interval around zero. But we see the derivative is positive for $|y|$ large, thus the function is not injective.
A mistake in your analysis is the absolute value in your expression $ \sqrt {\frac {|a|} 3}$. It should just be $ \sqrt {\frac {-a}3}$.
A: $$x^3+ax=y^3+ay\iff (x-y)(x^2+xy+y^2+a)=0$$
If the map is injective then the quadratic in $\;x\;,\;\;x^2+yx+y^2+a\;$ has to have negative discriminant:
$$y^2-4y^2-4a=-(3y^2+4a)<0\;\;,\;\;\text{for }\;\;a>0$$ 
so it must be $\;a>0\;$, otherwise we get a contradiction.
