Algebraic manipulation of a limit. What are the algebraic manipulations and steps that makes the limit
\begin{equation}
\lim_{x\to2}\left(\frac{x^3-8}{x-2}\right)
\end{equation}
equal to 
\begin{equation}
\lim_{x\to2}(x^2+2x+4)
\end{equation}
It's probably trivial, I just don't seem to be able to see it.

Based on @Kushal Bhuyan hint

Use $a^3-b^3=(a-b)(a^2+ab+b^2)$ formula

In our case 
$$\lim_{x\to2}\left(\frac{x^3-8}{x-2}\right)
=\lim_{x\to2}\left(\frac{x^3-2^3}{x-2}\right)
$$
Then $x^3-2^3=(x-2)(x^2+2x+2^2)$, therefore:
$$\require{cancel}\lim_{x\to2}\left(\frac{\cancel{(x-2)}(x^2+2x+2^2)}{\cancel{x-2}}\right)
=\lim_{x\to2}(x^2+2x+4)=2^2+2\cdot 2+4=\color{blue}{12}$$
