Proving that $\alpha:\mathbb{R}\to\mathbb{R}$ where $\alpha(x)=\frac{x^{3}}{x^{2}+1}$ is bijective Using the usual method that I was taught, in order to find that something is injective, I assume $f(x)=f(y)$ for some $x,y\in\mathbb{R}$, and then I demonstrate that $x=y$. This is not working out so well for me, using the following steps:
\begin{align}
\frac{x^{3}}{x^{2}+1}&=\frac{y^{3}}{y^{2}+1}\\
x^3(y^2+1)&=y^3(x^2+1)\\
x^3y^2+x^3&=x^2y^3+y^3\\
x^3-y^3&=x^2y^3-x^3y^2\\
(x-y)(x^2+xy+y^2)&=x^2y^2(y-x)\\
(x^2+xy+y^2)&=-x^2y^2\\
x^2+y^2&=-x^2y^2-xy\\
x^2+y^2&=-xy(xy+1)
\end{align}
And here is where I'm stuck. Am I assuming correctly that I should find that $x=y$ somehow?
 A: You're almost done proving that $f$ is injective.
If $f(x)=f(y)$, then $x$ and $y$ have the same sign, because the denominators are always positive. Hence, $xy\ge0$.
Now,
$
x^2+xy+y^2=-x^2y^2
$
implies
$
x^2+xy+y^2+x^2y^2=0
$.
The LHS is a sum of positive terms and so each term must be zero. This means that $x=y=0$.
This proves that $f$ is injective.
(Note that when you deduced $x^2+xy+y^2)=-x^2y^2$ from $(x-y)(x^2+xy+y^2)=x^2y^2(y-x)$ by canceling $x-y$, you assumed that $x-y\ne0$, that is, that $x\ne y$.)
To see that $f$ is surjective, take $a\in\mathbb R$. Then the equation $f(x)=a$ is a cubic equation in $x$ and so has a real solution.
A: Assuming that $f(x) = f(y)$ and showing that $x = y$ is a correct method to show that $f$ is injective. You have one subtle error in your work. 

$\begin{align} \frac{x^{3}}{x^{2}+1}&=\frac{y^{3}}{y^{2}+1}\\
 x^3(y^2+1)&=y^3(x^2+1)\\ x^3y^2+x^3&=x^2y^3+y^3\\
 x^3-y^3&=x^2y^3-x^3y^2\\ (x-y)(x^2+xy+y^2)&=x^2y^2(y-x)\\
 (x^2+xy+y^2)&=-x^2y^2\\ \end{align}$

At the last step shown above, you divided both sides by $x-y$, which could potentially be $0$. 
Continuing from above, you get that $f(x) = f(y)$ implies $$(x-y)(x^2+xy+y^2)=x^2y^2(y-x)$$, which can be rewritten as $$(x-y)(x^2y^2+x^2+xy+y^2) = 0$$
Here, there are two possibilities. Either $x-y = 0$, i.e. $x = y$ (which is what we want to show) or $x^2y^2+x^2+xy+y^2 = 0$. So you need to show that $x^2y^2+x^2+xy+y^2 = 0$ isn't possible unless the condition $x = y$ is also met. 
This is easy to show since we can write $x^2y^2+x^2+xy+y^2$ as $(xy)^2 + \dfrac{3}{4}(x+y)^2 + \dfrac{1}{4}(x-y)^2$, which can only be $0$ if $xy = x+y = x-y = 0$, i.e. $x = y = 0$. 
A: Here is an explicit inverse. 
$$x=\frac{\sqrt[3]{2y}}{\sqrt[3]{1+\sqrt{1+\frac{4}{27}y^2}}+\sqrt[3]{1-\sqrt{1+\frac{4}{27}y^2}}}$$
To find this, start with $y=\frac{x^3}{x^2+1}$, and solve for $x$ using Cardano's cubic solution. I subbed $z=x^{-1}$ to get a cubic that was already missing its quadratic term, which simplified the Cardano method a lot.
