How to prove $x^3$ is strictly increasing I am trying to use $f(x)=x^3$ as a counterexample to the following statement. 
If $f(x)$ is strictly increasing over $[a,b]$ then for any $x\in (a,b), f'(x)>0$. 
But how can I show that $f(x)=x^3$ 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. 
