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 $x^3 = (x+a)^3$.

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 $x=y=0$
