Proving Bijections i'm working on this question here but I am having some trouble:
$f: ℝ ⇒ ℝ$, $f(x) = x^3 - 6x$.
A) Is $f(x)$ injective?
B) Is $f(x)$ surjective?
C) Is $f(x)$ bijective?
My attempted solution:
A) $f(x)$ is injective if and only if $f(x) = f(y)$ $⇒$ $x = y$
$⇔$ $x^3 - 6x = y^3 - 6y$
$⇔$ $x(x^2 - 6) = y(y^2 - 6)$
$⇔$ $x(x - √6)(x + √6) = y(y - √6)(y + √6)$ I saw this step in a similar example but don't understand exactly what is going on here, an explanation would be great 
At this point I imagine from that last line above these $2$ values for x are derived from it?($0$ and $√6$)
$f(0) = x^3 - 6x$
$=$ $0^3 - 6(0) $
$= 0$ 
$f(√6) = x^3 - 6x$  
$= √6^3 - 6(√6)$
$= 0$
$∴ f(x)$ is not injective since $f(0) = 0 = f(√6)$
B) To show f(x) is surjective, we can assume $y =  x^3 - 6x$(a bit stuck on this step, or maybe there is nothing that can be done ∴ I know it's not surjective right away?)
$y =  x^3 - 6x $
what more can I do from here?
C) Regardless of figuring out part B or not, I still know that $f(x)$ is not a bijection since it's not injective.
Thanks!

Your argument for (A) is correct, but longer than necessary. All that you have to say is that $f(0) = f(\sqrt{6})$.
(C) is right too, for the reason you give.
For (B) - sketch the graph of the function. Can you see what happens as $x$ becomes large, or large (in absolute value) and negative?
(For future reference: sketching the graph would have been a good place to start thinking about all three questions.)
PS Please use mathjax to format the mathematics in any other questions or answers you want to post here.
