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The problem is asking you to completely factor the expression \( y^3 - y^2 - 16y + 16 \) by grouping.

### Step-by-Step Solution:
1. **Group the terms:** 
   \[
   (y^3 - y^2) - (16y - 16)
   \]

2. **Factor out the common factors in each group:**
   - For the first group \( y^3 - y^2 \), the common factor is \( y^2 \):
     \[
     y^2(y - 1)
     \]
   - For the second group \( -16y + 16 \), the common factor is \( -16 \):
     \[
     -16(y - 1)
     \]

   So the expression now is:
   \[
   y^2(y - 1) - 16(y - 1)
   \]

3. **Factor out the common binomial factor \( (y - 1) \):**
   \[
   (y - 1)(y^2 - 16)
   \]

4. **Factor the difference of squares in \( y^2 - 16 \):**
   - \( y^2 - 16 \) can be factored as \( (y - 4)(y + 4) \) since \( 16 = 4^2 \).

   Therefore, the fully factored expression is:
   \[
   (y - 1)(y - 4)(y + 4)
   \]

### Final Answer:
The expression \( y^3 - y^2 - 16y + 16 \) factors to:
\[
(y - 1)(y - 4)(y + 4)
\]

Do you want further details or have any questions?

Here are some related questions you might explore:
1. How do you factor cubic polynomials in general?
2. What are some common factoring techniques for polynomials?
3. How does recognizing the difference of squares help in factoring?
4. What is the significance of the common factor in polynomial grouping?
5. How can you verify if your factoring is correct?

**Tip:** Always check your factored expression by expanding it to ensure it matches the original polynomial.

Completely factor the expression by grouping if possible. y^3 - y^2 - 16y + 16

Learn how to completely factor the polynomial expression y^3 - y^2 - 16y + 16 using the grouping method. This problem covers essential algebraic concepts such as the difference of squares and common factoring, making it suitable for students in Grades 9-10.

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