How to factor $x^4-81$? So I was trying to evaluate the following limit:

$$f(x) = \lim \limits_{x\to 3} \frac{x^4 - 3x^3}{x^4-81}$$

To do so, you would need to factor the denominator to $(x^2+9)(x+3)(x-3)$ and cancel out $(x-3)$. If I wouldn't have seen it being factored this way, it would've been pretty difficult for me to factor the denominator in 3 terms. Is there a simple way/method to factor cases like that, that would work in most cases?

Thanks in advance!
Hopefully you know that $a^2-b^2=(a+b)(a-b)$

We can use this on the initial equation to get $$x^4-81=(x^2+9)(x^2-9)$$

We can then use the same technique on the second bracket to get $$x^2-9=(x+3)(x-3)$$

This gives us our final answer $$x^4-81=(x^2+9)(x+3)(x-3)$$
Write this as $\lim_{x\to 3} \frac{x^3(x-3)}{(x-3)(x+3)(x^2+9)}=\lim_{x\to 3} \frac{x^3}{(x+3)(x^2+9)}=\frac{27}{108}=\frac{1}{4}$