"Introducing something extra" In my textbook "Calculus, Concepts and Contexts" - by James Stewart
There is a section on problem solving methods called "Introducing something extra"
Given the problem: 
$$ \lim_{x \to 0} \frac{ \sqrt[3]{1+cx}-1}{x}$$ where $c$ is constant - Stewart suggests this method and introduces a new variable $t$, which is expressed
$$ t= \sqrt[3]{1+cx}$$
$$ x= \frac{t^3-1}{c} (\text{if }c\not=0)$$ 
and now suddenly 
$$ 
\lim_{x \to 0} \frac{ \sqrt[3]{1+cx}-1}{x} = 
\lim_{t \to 1} \frac{ t-1}{(t^3-1)/c} =
\lim_{t \to 1} \frac{c(t-1)}{t^3-1} 
$$
Factoring from this point is a simple difference of cubes approach, which I am familliar with.
However, I am really having trouble with the concept of introducing $t$ - It seems a fair number of steps were skipped in the text's explanation. 
Can anyone really break it down? Also is this technique normally called the substitution rule?
Probably my foremost concern is, why did we choose $t= \sqrt[3]{1+cx}$ 
When it's only one of the terms from the original numerator of the problem?

This is indeed normally called variable substitution. You want to take a limit at $0$ of this function:
$$f(x)= \frac{ \sqrt[3]{1+cx}-1}{x}$$
Now if we define a new function:
$$g(x)=\frac{x-1}{(x^3-1)/c}$$
Then we have $f(x)=g(\sqrt[3]{1+cx})$.
The idea now is that as $x$ approaches $0$, $\sqrt[3]{1+cx}$ approaches $1$, and so the limit as $x\to 0$ of $g(\sqrt[3]{1+cx})$ should simply be the limit as $x\to 1$ of $g(x)$. The exact conditions under which this works, and the proof that it does, are available in calculus textbooks or online.
