Showing that $f(x) = x^3$ is injective?

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 $x=y$.

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 $x=y=0$