Evaluate the limit as $x$ approaches $3$ : $\displaystyle\frac{x^3 -6x+2}{x^2+2x-3}$ Evaluate the limit x approaches 3 : $\displaystyle\frac{x^3 -6x+2}{x^2+2x-3}$
I am currently doing a limit problem, will be doing many in the coming months. I was wondering how do I factor a cube such as $x^3 -6x+2$? Is there a special rule? 
The ones I am familiar with are  $$a^3 + b^3 = (a + b)(a^2 – ab + b^2)$$ 
$$a^3 – b^3 = (a – b)(a^2 + ab + b^2)$$
but that does not work here. I also can not factor anything out.

If $f(x) = x^{3} -6x +2$ and $g(x) = x^{2} +2x-3$
You have $\lim\limits_{x \to a} \frac{f(x)}{g(x)}$ = $\frac{\lim\limits_{x \to a} f(x)}{\lim\limits_{x \to a} g(x)}$ with $a\in \mathbb{R}$ and obviously $g(x)\ne {0}$ when $x\rightarrow a$
Here $a=3$
so for $x \rightarrow 3$ :
$f(x) \rightarrow 11$
and $g(x) \rightarrow 12$
The final limit is : $\frac{11}{12}$
