prove $f(x)=x^3+x$ is one to one and onto The Problem
For my discrete structures course, I need to prove that $f(x)$ is one-to-one and onto, with $f:{\rm I\!R}\rightarrow{\rm I\!R}$ where $f(x)=x^3+x$. Based on the graph, this function is one-to-one and onto, and Wolfram confirms this, but I don't know how to approach the actual proof.
Proving it One-to-One
I understand a function $f(x)$ is one-to-one if for $x_1,x_2\in{\rm I\!R}$, if $f(x_1)=f(x_2)$ implies $x_1=x_2$.
 The problem is when I set $f(x_1)=f(x_2)$ this, I eventually get to
$$\sqrt[3]{x_1^3+x_1}=\sqrt[3]{x_2^3+x_2}$$
It's at this point I'm stuck, and don't know how to progress any further.
Proving it Onto
So I understand to prove a function is onto, you solve $f(x)$ for $y$, and use the result as input to $f(x)$, and if $f(x)=y$, the function is onto. But I run into a similar problems where
$$x^3+x=y$$
I could take a cubic root or rearrange the variables all I want but I can't think of a way to isolate $x$ and get $x=something$
 A: Let $f:\Bbb R\to\Bbb R$ be the function $f(x)=x^3+x$. Then $f'(x)=3x^2+1\ge 1>0$, so $f$ is strictly monotone, thus injective (one-to-one). Then the limits of $f$ at $\pm\infty$ are respectively $\pm\infty$, and from the continuity of $f$ each value in between is taken.
Note: One can also show algebraically that $f$ is injective, so assume $f(a)=f(b)$, then
$$
0=f(a)-f(b)=(a-b)\underbrace{(a^2+ab+b^2+1)}_{\ge 0+1>0}\ ,
$$
so the factor $(a-b)$ must vanish, so $a=b$.
A: Surjectivity:  For each $y$ polynomial equation $$x^3+x=y$$ is of odd degree, so it must have at least one real solution and thus $x^3+x$ is surjective.
Injectivety: Say there are $a$, $b$ such that $f(a)= f(b)$ and suppose $a\ne b$, then $$ (a-b)(a^2+ab+b^2+1)=0\implies a^2+ab+b^2+1=0$$
so multilpying this with 2 we get $$ (a+b)^2+a^2+b^2 +2 =0$$ which is clearly nosense. So $a=b$. 
A: Partial answer:
1) Injective:
$x_1,x_2 \in \mathbb{R}$, and let
$f(x_1)=f(x_2).$
$x_1^3 +x_1=x_2^3+x_2$;
$x_1^3-x_2^3 +x_1-x_2=0;$
$(x_1-x_2)(x_1^2+x_1x_2+x_2) +(x_1-x_2)=0;$
$(x_1-x_2)(x_1^2+x_1x_2+x_2^2+1)=0;$
It follows $x_1=x_2$ , and we are done,
since $(x_1^2+x_1x_2+x_2^2+ 1) >0$.
Recall: $a^2+b^2 \ge 2|ab|$,
$a^2+ab+b^2 \ge 2|ab| +ab \ge$
$|ab| \ge 0.$
$a^2+ab+b^2 +1 \ge 1 >0.$
2) $y=x^3+x$ , a polynomial of degree $3$ has at least one real root.(Cf. Answer by Maria Mazur).
For given $y$, it has exactly one real root since $y=f(x)$ is injective.
A: It is one-to-one because it is strictly increasing ($x>y\implies f(x)>f(y)$) and it is surjective be cause $\lim_{x\to\pm\infty}f(x)=\pm\infty$ and by the intermediate value theorem.
A: Just to take a different approach, you can use Cardano's formula: write the equation $x^3+x=y$ as $x^3+x-y=0$. The discriminant is
$$
\frac{y^2}{4}+\frac{1}{27}>0
$$
so the equation has a single real solution, for every $y$. This proves at once both injectivity and surjectivity.
You can even solve for the inverse function:
$$
x=\sqrt[3]{\frac{y}{2}+\sqrt{\frac{y^2}{4}+\frac{1}{27}}}+
  \sqrt[3]{\frac{y}{2}-\sqrt{\frac{y^2}{4}+\frac{1}{27}}}
$$
