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. 

Regarding (a), when you say "take cube root of both sides" you are (at least implicitly) assuming that the function is injective -- if it were not, the cube root would not be a well-defined operation.  Depending on what you are / are not allowed to assume, that could be regarded as a circular argument.
However, there is another approach.  Beginning with the assumption that $x^3=y^3$, you can manipulate the expression as follows:
$$x^3-y^3=0$$
$$(x-y)(x^2+xy+y^2)=0$$
Now if you can find a convincing argument that the second factor, $x^2+xy+y^2$, is always positive for any choice of $x$ and $y$, then it follows that the only way $x^3=y^3$ can be true is if $x=y$.
The factorization $x^3-y^3=(x-y)(x^2+xy+y^2)$, by the way, could also give you some insight into how to handle (b).
