Prove that $f(x)=x^3−x$ is surjective with elementary knowledge So I'm reading a book which requires no prior knowledge other than highschool algebra. I'm required to prove that the function from $\mathbb{R}$ to $\mathbb{R}$ $f(x) = x^3-x $ is a surjection, basically just using the definition of surjection and I'm guessing some clever manipulation that I haven't been able to come up with. So using calculus knowledge like in this question is not allowed. Any ideas on how to proceed?
 A: By definition of a surjective function the question statement is equivalent to:

Let $P(x)=x^3-x+c$, where $c\in\mathbb R$, then prove that the polynomial $P(x)$ always has at least one real root.


Suppose that $x_1,x_2,x_3$ are roots of the polynomial  $P(x)$, such that $x_1,x_2,x_3\in\mathbb C\setminus \mathbb R$.
Let $x_1=a+bi$, where $a\in\mathbb R, b\in \mathbb R\setminus \{0\}$.
We have,
$$P(a{\color{#c00}{+}}bi)={\color{#c00}{ib}}(3a^2-b^2-1)\\
+\left(a^3-a(3b^2+1)+c\right)=0$$
This implies that, $3a^2-b^2-1=0$ and $a^3-a(3b^2+1)+c=0$.
This leads to,
$$P(a{\color{#c00}{-}}bi)={\color{#c00}{-ib}}(3a^2-b^2-1)\\
+\left(a^3-a(3b^2+1)+c\right)=0.$$

Thus, we have shown that, if $x_1=a+bi$ is one of the roots of the polynomial $P(x)=x^3-x+c$, then $x_2=a-bi$ is also one of the roots of $P(x)$.

Finally, using Vieta's formulas we observe that:
$$
\begin{align}&x_1+x_2+x_3=0\\
\implies &x_1+x_2=-x_3\\
\implies &x_3=-2a\in\mathbb R.\end{align}
$$
which contradicts the assumption $x_1,x_2,x_3\not\in\mathbb R$. Therefore, there exist at least one $x_i$,  such that $x_i\in\mathbb R.$
This completes the proof.
