Prove that $f(x) = x^3 -x $ is NOT Injective Okay, so I know there are plenty of injective/surjective (and thus, bijective) questions out there but I'm still not happy with the rigor of what I have done. So I'd really appreciate some help! 
So the question actually asks me to do two things:
(a) give an example of a cubic function that is bijective. Explain why it is bijective.
(b) give an example of a cubic function that is not bijective. Explain why it is not bijective. 
So for (a) I'm fairly happy with what I've done (I think):
$$ f: \mathbb R \rightarrow \mathbb R , f(x) = x^3$$
So we know that to prove if a function is bijective, we must prove it is both injective and surjective. 
Proof: $f$ is injective
Let: $$x,y \in \mathbb R : f(x) = f(y)$$ 
$$x^3 = y^3$$            (take cube root of both sides)
$$x=y$$ 
Proof: $f$ is surjective
Let: $$y \in \mathbb R$$
$$x = \sqrt[3]{y}$$
$$f(x) = (\sqrt[3]{y})^3 = y$$
So I believe that is enough to prove bijectivity for $f(x) = x^3$. Keep in mind I have cut out some of the formalities i.e. invoking definitions and sentences explaining steps to save readers time. This is just 'bare essentials'.
So for (b)
$$f: \mathbb R \rightarrow \mathbb R , f(x) = x^3 - x$$
Now I'm just going to try and prove it is NOT injective, as that should be sufficient to prove it is NOT bijective. 
Let:
$$x,y \in \mathbb R : f(x) = f(y)$$
$$x^3 - x = y^3 - y$$
This is about as far as I get. Send help. Thanks everyone. 

Injectivity:
A function $f$ from $X\to Y$ is said to be injective iff the following statement holds true:

for every $x_1,x_2\in X$ if $x_1\neq x_2$ then $f(x_1)\neq f(x_2)$

The negation of this then yields:
A function $f$ from $X\to Y$ is not injective iff there exists $x_1,x_2\in X$ such that $x_1\neq x_2$ but $f(x_1)=f(x_2)$
In the case of the cubic in question, it is an easily factorable polynomial and we can find multiple distinct roots.  $f(x)=x^3-x=x(x^2-1)=x(x+1)(x-1)$
We know that a root of a polynomial is a number $\alpha$ such that $f(\alpha)=0$

 So, $f(1)=f(0)=f(-1)=0$ despite $1,0,-1$ all being distinct unequal numbers in the domain.  This shows that it is not injective, and thus not bijective.


As an aside, one can prove that any odd degree polynomial from $\Bbb R\to \Bbb R$ must be surjective by the fact that polynomials are continuous and the intermediate value theorem.
