Can someone walk me through this differential problem? I'm having a little difficulty understanding how to do the .05 using differentials. I'm just hoping someone can walk me through, step by step, and explain why they are during it that way.
$$\sqrt[3] {27.05}$$
Edit
Apologies to all, I wasn't very clear on what I was asking. 
I was looking for someone to explain how to find an approximate answer to the equation above using differentials. (such as @Arturo has done)
@Sasha @Ross  I apologize for wasting your time when answering the question using series. 

The general philosophy when using differentials:

You are trying to find the value of a function at a difficult point, but there is a point, very close to the difficult point, where the function is easy to evaluate. Then one can use the formula
  $$f(a+\Delta x) = f(a)+\Delta y \approx f(a)+dy = f(a)+f'(a)dx=f(a)+f'(a)(x-a)\quad
\text{ for }x\text{ near }a,$$
  where $\Delta x = dx$ is the change in $x$ (how far the difficult point is from $a$, which is $x-a$), $\Delta y$ is the change in value (which is $f(x)-f(a)$), and $dy$ is the differential of $y$ (which is $dy = f'(x)dx$). If $\Delta x=dx$ is small, then $\Delta y$ is very close to $dy$, so we can use $dy$ to approximate.

So you want to do the following:


*

*Identify the function you are trying to evaluate. This will be your $f$.

*Find a point near the point where you are trying to evaluate where the function is easy: this will be your $a$.

*Compute $f(a)$ (which is supposed to be easy) and compute $f'(a)$ (which hopefully will be easy as well).

*Plug into the formula.
In this case, $f(x) = \sqrt[3]{x}$, and the "hard point" is $x=27.05$. Is there a point $a$ near $x$ where $f$ is easy to evaluate? Yes! $f(27)$ is easy. So take $a=27$, $\Delta x = dx = .05$, and use the approximation. You'll have to compute $f'(x)$ and $f'(27)$.
