<s>[INST] 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?
 [/INST] 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.
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