Can't understand proof for why $f(x) =x^3$, $f: \mathbb{R} \rightarrow \mathbb{R}$ is injective Edit: This is the proof as written in a lecture, I did not come up with this, and it's possible it's entirely nonsensical.
Here's the proof:
Let $y \in \mathbb{R}$ and $x \in \mathbb{R}$ such that:
$$x=\sqrt[3]{y}$$
Therefore:
$$f(x)\leq x^3 = y$$
That's it. Now, clearly the function is injective because I know that this sort of power function has no same value for different values of $x$, but I don't see how this proof shows it. 
 A: A map $f:A\to B$ is, by definition, injective if $f(a)=f(b)\Rightarrow a=b$ for all $a,b\in A$. In this case, we have to prove $a^3=b^3\Rightarrow a=b$ for all $a,b\in\mathbb{R}$. Well, suppose that $a=b+c$ for some $c\in\mathbb{R}$. Then, we get $$(b+c)^3=b^3$$ so $$3b^2c+3bc^2+c^3=0$$ so either $c=0$ or $$3b^2+3bc+c^2=0.$$ This equation is quadratic in $c^2$, and the discriminant is $(3b)^2-4\cdot1\cdot3b^2=-3b^2\leq0$. If $b\neq 0$ then this does not have any solutions so in that case $c=0$ meaning that $a=b$. If $b=0$ then we use the middle equation to deduce that $c^3=0$ hence $c=0$ as well. In all cases, $c=0$ so $a=b$.
A: I don't completely understand the proof you are trying, but here's another way:
Let $f(x) = f(y)$ $\implies x^3 = y^3  \implies x^3 - y^3 = 0 \implies (x-y)(x^2 + xy + y^2) = 0 $
Now, $ (x^2 + xy + y^2) = \frac{1}{2} [(x+y)^2 + x^2 + y^2]$. So, $ (x^2 + xy + y^2) = 0$ iff $ x = y = 0$ So, in this case, $ x = y$
If $ (x^2 + xy + y^2) \neq 0$, then $ (x-y) = 0 \implies x = y $. Hence, $f$ is injective.
