How would I solve this equation? $x_1^3 - x_1 = x_2^3 - x_2$

Question

I want to find whether the function $h(x)=x^3-x$ is injective or not. If $\forall x_1, x_2 \in \mathbb R, f(x_1) = f(x_2) \Rightarrow x_1 = x_2$, then how do I solve this $x_1^3 - x_1 = x_2^3 - x_2$?

Answer 1

$$x_1^3 - x_1 = x_2^3 - x_2 $$

$$(x_1-x_2)(x_1^2+x_1x_2+x_2^2-1)=0$$

$$ x_1=x_2 $$ or $$x_1^2+x_1x_2+x_2^2-1=0$$

As you notice, $x_1=1$ and $x_2=-1$ satisfy the $$x_1^2+x_1x_2+x_2^2-1=0$$

Answer 2

$h(x)=x^3-x$ is not injective because $h(1)=h(-1) =0$

Answer 3

The cubic in question is an easily factorable polynomial and we can find multiple distinct roots.

$f(x)=x^3-x=x(x^2-1)=x(x+1)(x-1)$

So, $f(1)=f(0)=f(−1)=0 $ despite $1,0,−1$ all being distinct unequal numbers in the domain. This shows that it is not injective.