The given limit expression is:

\[
\lim_{{x \to 2}} \frac{{x^3 - 8}}{{3x^2 - 6x}}
\]

To solve this, we can follow these steps:

### Step 1: Check for direct substitution
First, try substituting \(x = 2\) into the expression:

\[
\frac{{2^3 - 8}}{{3(2^2) - 6(2)}} = \frac{{8 - 8}}{{3(4) - 12}} = \frac{0}{0}
\]

Since this results in an indeterminate form \( \frac{0}{0} \), we need to simplify the expression further.

### Step 2: Factor the numerator and denominator

- The numerator \(x^3 - 8\) is a difference of cubes. This can be factored as:

\[
x^3 - 8 = (x - 2)(x^2 + 2x + 4)
\]

- The denominator \(3x^2 - 6x\) can be factored as:

\[
3x^2 - 6x = 3x(x - 2)
\]

### Step 3: Simplify the expression
Now substitute the factored forms into the original limit:

\[
\lim_{{x \to 2}} \frac{{(x - 2)(x^2 + 2x + 4)}}{{3x(x - 2)}}
\]

We can cancel the \( (x - 2) \) terms from the numerator and denominator, assuming \( x \neq 2 \):

\[
\lim_{{x \to 2}} \frac{{x^2 + 2x + 4}}{{3x}}
\]

### Step 4: Apply direct substitution again
Now, substitute \(x = 2\) into the simplified expression:

\[
\frac{{2^2 + 2(2) + 4}}{{3(2)}} = \frac{{4 + 4 + 4}}{{6}} = \frac{{12}}{{6}} = 2
\]

### Final Answer:
\[
\lim_{{x \to 2}} \frac{{x^3 - 8}}{{3x^2 - 6x}} = 2
\]

Would you like more details or further clarification?

Here are some additional questions related to this concept:
1. How do you solve limits involving indeterminate forms other than \(\frac{0}{0}\)?
2. What is the factorization process for cubes and squares in polynomials?
3. Can you explain why cancelling terms like \( (x - 2) \) is valid?
4. How do you apply L'Hopital's Rule to solve limits?
5. What are common indeterminate forms in limits, and how can we simplify them?

**Tip:** Always check for factorization possibilities when you encounter indeterminate forms like \(\frac{0}{0}\).

Explore how to evaluate the limit lim x→2 (x^3−8) / (3x^2−6x) using factorization techniques. This detailed solution covers key calculus concepts, including limits, indeterminate forms, and the factorization of cubes. Suitable for students in Grades 11-12.

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