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The given expression is:

\[
\frac{y^2 - 9}{y^2 - 6y + 9}
\]

### Step 1: Factor both the numerator and the denominator.

#### Numerator:
The numerator is \( y^2 - 9 \), which is a difference of squares:
\[
y^2 - 9 = (y - 3)(y + 3)
\]

#### Denominator:
The denominator is \( y^2 - 6y + 9 \), which is a perfect square trinomial:
\[
y^2 - 6y + 9 = (y - 3)^2
\]

### Step 2: Simplify the expression.

Now that we have factored both the numerator and the denominator, the expression becomes:

\[
\frac{(y - 3)(y + 3)}{(y - 3)^2}
\]

We can cancel out one factor of \( y - 3 \) from both the numerator and denominator (as long as \( y \neq 3 \) to avoid division by zero):

\[
\frac{y + 3}{y - 3}
\]

### Step 3: State the domain.

The domain is determined by the values of \( y \) that do not make the denominator zero. In the original expression, the denominator is \( y^2 - 6y + 9 = (y - 3)^2 \). This expression is zero when \( y = 3 \). Therefore, \( y = 3 \) is excluded from the domain.

Thus, the domain is:
\[
y \in \mathbb{R}, \quad y \neq 3
\]

### Final Answers:
- **Domain of original expression**: \( y \in \mathbb{R}, y \neq 3 \)
- **Simplified expression**: \( \frac{y + 3}{y - 3} \)

Would you like a more detailed explanation or have any questions?

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Here are some related questions to expand on this topic:

1. What happens when you have complex numbers in a similar expression?
2. How do you simplify expressions involving cube differences, such as \( x^3 - 8 \)?
3. What is the procedure for factoring trinomials where the leading coefficient is not 1?
4. Can you explain how to find the domain of more complex rational functions?
5. How do you approach rational expressions with higher powers?

**Tip:** Always check for restrictions on the domain by identifying values that would make any denominator zero.

Simplify and state the domain. 
(y^2 - 9) / (y^2 - 6y + 9)

Simplify the rational expression (y^2 - 9) / (y^2 - 6y + 9) by factoring and canceling common factors. This exercise includes factoring techniques such as the difference of squares and perfect square trinomials, and determining the domain where the expression is defined. Suitable for students in Grades 9-11.

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