Showing that $f(x) = x^3$ is injective? This is my attempt. Is it right? Is there a simpler way? Is there a way which relies on less background knowledge?

A function is injective iff $f(x) = f(y) \implies x = y$. This is equivalent to $x \not = y \implies f(x) \not = f(y)$. We will prove this latter statement by contradiction:
Suppose there are two numbers $x$, $x+a$ with $a >0$ and $x \not = x+a$ but $x^3 = (x+a)^3$. Expanding the RHS, subtracting $x^3$, and dividing both sides by $a>0$ yields $0 = 3x^2 + 3xa + a^2$. Taking this last equation as a quadratic in $x$, it has a real solution iff the discriminant is non-negative. But the discriminant is $9a^2 - 12a^2 = -3a^2$, so there are no real $x$ which satisfy $x^3 = (x+a)^3$.
 A: I might say $a\ne 0$ instead of $a>0$ just for a nice symmetry, but that's purely aesthetic, your answer is a good one. An alternative is let $x\ne y$ and consider

$$x^3-y^3 = (x-y)(x^2+xy+y^2)$$

Then by the AM-GM inequality, $x^2+y^2 > 2|xy|>|xy|$--strict because $x\ne y$--so the only way for this difference to be $0$ is if $x=y$.
A: Suppose $$x^3=y^3$$ then $$(x-y)(y^2+xy+x^2)=0$$ so either $x=y$ or $$x^2+xy+y^2=\left(x+\frac y2\right)^2+\frac {3y^2}4=0$$
But this is strictly non-negative and is zero only if $x=y=0$
A: Is there a way which relies on less background knowledge?
Yes, there is (probably the most elementary one, which is based essentially in the monotonicity of multiplication):

Claim 1. If $x\neq 0$, $y\neq 0$ and $x^3=y^3$ then $x$ and $y$ have the same sign.

Proof: Assume that $x^3=y^3$ with $x>0$ and $y<0$. Then
$$0<x^3=y^3=(-1)^3(-y)^3<0,$$
which is a contradiction. $\square$

Claim 2: $f(x)=x^3$ is injective.

Proof: Note that $x^3=xxx\neq 0$ for all $x\neq 0$ and $0^3=0$. So, if $f(x)=x^3$ is not injective then there exist two real numbers $x$ and $y$ such that:


*

*$x\neq 0$, $y\neq0$

*$x\neq y$

*$x^3=y^3$
Without loss of generality, we can assume that $x<y$. It follows from Claim 1 that $0<x<y$ or $0>y>x$. Thus
$$x^3=xxx<yyy=y^3$$
which is a contradiction. $\square$
Remark. Note that this proof relies only in the following elementary facts:


*

*$a<b \text{ and }c>0\quad \Longrightarrow \quad ac<bc$

*$a<b \text{ and }c<0\quad \Longrightarrow \quad ac>bc$

*$ab=0\quad \Longrightarrow \quad a=0\text{ or }b=0$

*$-a=(-1)a\text{ and } -(-a)=a$

*$ab=ba$ 

*$0^2=0$

*$(-1)^2=1$

*$-1<0$

