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.

There is no need to factor numerator or denominator because at $x\to3$ numerator and denominator are defined and denominator isn't $0$, so just write:
$$\lim_{x\to3}\frac{x^3-6x+2}{x^2+2x-3}=\frac{3^3-6\cdot3+2}{3^2+2\cdot3-3}=\frac{11}{12}$$
