Prove that $|\sin x-x |\le \frac{1}{6}|x|^3$ Prove that $|\sin x-x |\le \frac{1}{6}|x|^3$
I've been trying to solve this problem for the last hour but I just can't crack it. I know I'm supposed to use Cauchy's Mean Value Theorem somehow but can't figure out how.
Edit: This problem is from chapter 4 of Foundations of Analysis by Taylor. The book hasn't covered Taylor series yet. I am supposed solve this using a method that is already covered. It is very likely that the intended solution involves the usage of Cauchy's form of the mean value theorem.

A solution using Cauchy's Mean-Value Theorem: With 
$$
 f(x) = \sin(x) - x, g(x) = x^3
$$
we have for all $x > 0$
$$
 \frac{\sin(x)-x}{x^3} = \frac{f(x)-f(0)}{g(x)-g(0)} = \frac{f'(c)}{g'(c)} 
= \frac{\cos(c)-1}{3c^2}
$$
for some $c$ between $0$ and $x$. Now apply Cauchy's Mean-Value Theorem again twice to conclude that this is
$$
\begin{align}
 &= \frac{-\sin(d)}{6d} \quad \text{for some $d$ between $0$ and $c$} \\
 &= \frac{-\cos(e)}{6} \quad \text{for  some $e$ between $0$ and $d$} 
\end{align}
$$
and therefore
$$
\left| \frac{\sin(x)-x}{x^3} \right| = \frac{|\cos(e)|}{6} \le \frac 16 \, .
$$
