I'm having some trouble with this question

Use the mean value theorem to show that:

$\displaystyle

1)\ \frac{4}{27}<3-23^ \frac{1}{3}< \frac{3}{16}

$

Would you start of like this:

$\displaystyle \frac{f(27)-f(23)}{27-23}= f'(c),\ c \in (23,27)$

$\displaystyle Let\ f(x)=x^ \frac{1}{3}$

$\displaystyle \frac{27^ \frac{1}{3} - 23^ \frac{1}{3}}{27-23} = \frac{1}{3c^ \frac{2}{3}}$

$\displaystyle \frac{3- 23^ \frac{1}{3}}{4}= \frac{1}{3c \frac{2}{3}}$

$\displaystyle 23<c<27$

$\displaystyle 23^ \frac{2}{3}<c^ \frac{2}{3}< 27^ \frac{2}{3}$

$\displaystyle 3* 23^ \frac{2}{3}<3c^ \frac{2}{3}< 3*9$

$\displaystyle \frac{1}{3* 23^ \frac{2}{3}}> \frac{1}{3c^ \frac{2}{3}}> \frac{1}{27}$

From here on, I don't know what to do?