Applying the Taylor's Theorem, show that if $x > 0$ then $|(1+x)^{1/3} - (1 + \frac x3 - {x^2\over 9})| \le {5x^3\over 81}$ I've been struggling with this problem for weeks now. Seeing as the Taylor's Theorem has up to n terms, I have no idea where to start or stop with it as I wasn't given a n to work with. Can I use the MVT? How would I use the MVT? I'm truly lost, and any help would be appreciated.
Let $f(x)=(1+x)^{1/3}$. It is easy to verify that that the sum of the first three terms of the Taylor  expansion of $f(x)$ about $x=0$ is $1+\frac{x}{3}-\frac{x^2}{9}$.

Note that $f'''(x)=-\frac{10}{27}(1+x)^{-8/3}$.

By the Lagrange form of the remainder, the absolute value of the error when we truncate at the $x^2$ term in the Taylor expansion is
$$\frac{1}{3!}|f'''(c)| |x|^3,$$
where $c$ is between $0$ and $x$. In our case, that shows that the absolute value of the error is
$$\frac{1}{3!}\cdot \frac{10}{27}(1+c)^{-8/3}x^3.\tag{1}$$
Since $x$ is positive, $0\lt (1+c)^{-8/3}\lt 1$. The desired result now follows from (1).

Remark: The Lagrange form of the remainder can be proved using nothing more than the Mean Value Theorem. Probably, however, you are expected to just use the Lagrange form of the remainder, and not to prove it, even for this particular case.
Hint: Check the remainder of alternating series. 

Added: In your case $f(x) = (1+x)^{1/3}$


  $$\Bigg| (1+x)^{1/3}-(1-x/3+x^2/9)\Bigg|\leq a_3$$


Where 


  $$a_3=\frac{f^{(3)}(0)} {3!}x^3. $$


If calculate the third derivative of the function at $x=0$ you will get


  $$f^{(3)}(0)=\frac{10}{27}. $$