How to prove $x^3$ is strictly increasing

Consider $a^3-b^3$, where $a\gt b$. We want to show this is positive. We have $$a^3-b^3=(a-b)(a^2+ab+b^2).$$ Note that $a^2+ab+b^2$ is positive unless $a=b=0$. There are many ways to do this. For example, $$a^2+ab+b^2=\frac{1}{2}\left(a^2+b^2+(a+b)^2\right).$$ More conventionally, complete the square. We have $4(a^2+ab+b^2)=(2a+b)^2+3b^2$.

Remark: If we are in a calculus mood, note that $$\frac{a^3-b^3}{a-b} =3c^2$$ for some $c$ between $a$ and $b$. This argument breaks down if $c=0$. But that can only happen when $0$ is in the interval $(b,a)$. That means $b$ is negative and $a$ is positive, making the inequality $a^3\gt b^3$ obvious.

Well, you know that $f$ is strictly increasing in $[0,+\infty)$ and in $(-\infty,0]$. Morevoer, $f$ is positive in the first interval and negative in the second one!