The given expression is:

$\lim_{{x \to 3}} \frac{x^3 - 27}{x - 3}$

Let's break it down:

1. First, notice that $x^3 - 27$ is a difference of cubes, which can be factored as:

$x^3 - 27 = (x - 3)(x^2 + 3x + 9)$

2. Now substitute this into the original limit expression:

$\lim_{{x \to 3}} \frac{(x - 3)(x^2 + 3x + 9)}{x - 3}$

3. The $x - 3$ terms cancel out:

$\lim_{{x \to 3}} (x^2 + 3x + 9)$

4. Now, substitute $x = 3$ into the simplified expression:

$3^2 + 3(3) + 9 = 9 + 9 + 9 = 27$

Thus, the value of the limit is:

$\boxed{27}$

Would you like more details on factoring or limits?

Here are 5 related questions to explore:

1. How do you factor higher-degree polynomials in limits?
2. Can L'Hopital's rule be used for this limit problem?
3. What are other methods to solve limits involving indeterminate forms?
4. How does the difference of cubes factoring formula work?
5. Can this problem be solved graphically?

Tip: Always check for common factors when dealing with indeterminate forms in limits.