How to factor out third degree cubic equations while only having x^3 and a constant?

Question

Determine $$\lim_{x\to3}\frac{x^2-5x+6}{x^3-27}$$

I know how to factor cubic equations if they are of the form $ax^3+bx^2+cx+d=0$, but what happens if you only have the $ax^3$ and the $d$, as in $x^3 -27$? A good explanation would be appreciated.

Answer 1

There is a general identity: $$ a^3 - b^3 = (a - b) (a^2 + ab + b^2) $$

Answer 2

use that $$x^3-27=(x-3) \left(x^2+3 x+9\right)$$ after the formula $$a^3-b^3=(a-b)(a^2+ab+b^2)$$

Answer 3

The cubic $x^3 - c$ (with $ c \ne 0$) has one real root, the cube root $r$ of $c$. So over the real numbers it factors as $$ x^3 - c = (x-r)(x^2 + rx + r^2) $$ and two complex roots, $\omega r$ and $\omega^2 r$, where $\omega = (-1 + i\sqrt{3})/2$. You can find those with the quadratic formula for the second factor.

To find the limit in your question (added after my answer) you don't need to know anything about the complex roots.