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?

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.
