The limit you're trying to evaluate is:

$\lim_{{x \to 1}} \frac{x^3 - 1}{x - 1}$

If we substitute $x = 1$ directly, both the numerator and denominator become 0, resulting in an indeterminate form $\frac{0}{0}$. To resolve this, we can factor the numerator.

Step 1: Factor the numerator

We can factor $x^3 - 1$ as a difference of cubes:

$x^3 - 1 = (x - 1)(x^2 + x + 1)$

Step 2: Simplify the expression

Now, substitute the factored form of the numerator into the original limit:

$\frac{x^3 - 1}{x - 1} = \frac{(x - 1)(x^2 + x + 1)}{x - 1}$

Since $x - 1$ appears in both the numerator and denominator, we can cancel them out (as long as $x \neq 1$):

$= x^2 + x + 1$

Step 3: Evaluate the limit

Now, we can directly substitute $x = 1$ into the simplified expression:

$1^2 + 1 + 1 = 3$

Thus, the limit is:

$\lim_{{x \to 1}} \frac{x^3 - 1}{x - 1} = 3$

Let me know if you need more details or have any questions!

Here are 5 related questions to expand your understanding:

1. How would you compute $\lim_{{x \to 2}} \frac{x^3 - 8}{x - 2}$?
2. What is the general formula for factoring a difference of cubes?
3. Can you apply L'Hopital's Rule to evaluate $\lim_{{x \to 1}} \frac{x^3 - 1}{x - 1}$?
4. What happens if you apply direct substitution to limits involving indeterminate forms like $\frac{0}{0}$?
5. How does the factorization technique change for higher powers of $x$, like $x^4 - 1$?

Tip: When you encounter a limit with an indeterminate form $\frac{0}{0}$, try factoring or using L'Hopital's Rule to resolve it.