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!
 A: For proving surjectivity, note that $\lim\limits_{x\to\infty}f(x)=\infty$ and $\lim\limits_{x\to -\infty}f(x)=-\infty$.  Further note that $f(x)$ is a polynomial and it is known that polynomials are continuous.  An application of the Intermediate-Value-Theorem will show that for each possible value of $y$ it can be achieved by some $x$.
More specifically, we know that cubics (being polynomials of odd degree) will always have at least one real root.
Given a specific $y$, we try to find what value of $x$ will give that value of $y$.
$y=f(x)=x^3-6x\Rightarrow 0=x^3-6x-y$ where $y$ is known.  We know it will have a real root for $x$.  We do not necessarily need to find what exactly it is to be satisfied, but via cardano's method we find that it would be something along the lines of $\frac{\sqrt[3]{\sqrt{y^2-32}+y}}{\sqrt[3]{2}}+\frac{2\sqrt[3]{2}}{\sqrt[3]{\sqrt{y^2-32}+y}}$.  You are not expected to know this.

For proving/disproving injectivity, all you need to do is find an instance where $f(x)=f(y)$ despite $x\neq y$ (to prove it is not injective) or you need to show that $f(x)=f(y)$ implies that $x=y$.
With $f$ a polynomial of low degree, we attempt to factor it.  A polynomial is in factored form if it is of the form $f(x)=(x-\alpha_1)(x-\alpha_2)(x-\alpha_3)\cdots (x-\alpha_n)$.  Each of these terms $\alpha_i$ are what are called "roots" and are the points for which $f(x)=0$.
In your case $f(x)=x^3-x=(x-0)(x-\sqrt{6})(x-(-\sqrt{6}))$ so there are three roots.  $f(0)=f(\sqrt{6})=f(-\sqrt{6})=0$.  In particular, these roots are distinct, so we have found an occurrence of an $x$ and a $y$ where $f(x)=f(y)$ but $x\neq y$.
A: 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.
A: $$\begin{align}
  & f({{x}_{1}})=f({{x}_{2}})\Rightarrow {{x}_{1}}^{3}-6{{x}_{1}}={{x}_{2}}^{3}-6{{x}_{2}}\Rightarrow (\,{{x}_{1}}^{3}-{{x}_{2}}^{3})+(-6{{x}_{1}}+6{{x}_{2}})=0 \\ 
 & ({{x}_{1}}-{{x}_{2}})({{x}_{1}}^{2}+{{x}_{1}}{{x}_{2}}+x_{2}^{2})-6({{x}_{1}}-{{x}_{2}})=({{x}_{1}}-{{x}_{2}})({{x}_{1}}^{2}+{{x}_{1}}{{x}_{2}}+x_{2}^{2}-6)=0\, \\ 
 & \large\left\{ \begin{matrix}
   {{x}_{1}}={{x}_{2}}  \\
   {{x}_{1}}^{2}+{{x}_{1}}{{x}_{2}}+x_{2}^{2}-6=0\,\,\,\,\,\,\Rightarrow \,\,\,\,{{x}_{1}}=\frac{-{{x}_{2}}\pm \sqrt{-3{{x}_{2}}^{2}+24}}{2}  \\
\end{matrix} \right. \\ 
\end{align}$$
$f$ is surjective:
let $t\in\mathbb{R}$. We have $x^3-6x=t$. This equation has a real solution, because it's degree is odd.
