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.
 A: Use $a^3-b^3=(a-b)(a^2+ab+b^2)$ formula
A: 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:
$$\require{cancel}\lim_{x\to2}\left(\frac{\cancel{(x-2)}(x^2+2x+2^2)}{\cancel{x-2}}\right)
=\lim_{x\to2}(x^2+2x+4)=2^2+2\cdot 2+4=\color{blue}{12}$$
A: $$\dfrac{x^3-8}{x-2}$$
$$x^3-8\Rightarrow(x-2)(x^2+2x+4)$$
$$=\dfrac{(x-2)(x^2+2x+4)}{x-2}$$
$$=x^2+2x+4$$
A: Another way to solve this without knowing $a^3-b^3=(a-b)(a^2+ab+b^2)$. 
You can always apply polynomial long division in such fractions:
$$(x^3-8):(x-2)=x^2+2x+4$$ 
A: Assuming you are ignorant of the factorization formula for the difference of two cubes, you have to rediscover it.
From the problem statement, you know that $x^3-8$ has a root at $x=2$. Then the binomial will factor as
$$x^3-8=(x-2)P_2(x)$$ where $P_2$ is a quadratic trinomial. We can obtain it by long division
$$\begin{matrix}&1&0&0&-8\\2|&&2&4&8\\&1&2&4&0\end{matrix},$$ giving
$$x^3-8=(x-2)(x^2+2x+4).$$
The limit easily follows as $2^2+2\cdot2+4$.
