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?

You want to show that the function $f(x) = x^3$ is strictly increasing on $\mathbb{R}$.
Maybe you can just use the definition. That is, let $a < b$. Assume that $0<a$ (you can do the other cases I am sure). Let $h = b - a < 0$. You want to show that $f(a) < f(b)$.
So
$$\begin{align}
f(a) = a^3 &= (b - h)^3\\ &= b^3 - 3b^2h +3 bh^2 - h^3 \\
 &<b^3 \\ &= f(b).
\end{align}
$$
This is clear since $h > 0$, so $-3b^2h + 3bh^2 = -3hb(b - h) = -3hba < 0$.

All that said, you could, of course just use $f(x) = x$ as a counter example.
