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.

Discover how to solve the limit lim (x -> 1) (x^3 - 1)/(x - 1) by factoring the cubic expression and simplifying indeterminate forms. This problem covers calculus concepts and is suitable for high school and college students.

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