Solve the equation.

y cubed plus y squared equals 9 y plus 9y3+y2=9y+9

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Part 1

The solution set is 

StartSet nothing EndSetenter your response here.

​(Simplify your answer. Type an exact​ answer, using radicals as needed. Express complex numbers in terms of 

ii. 

Use a comma to separate answers as needed. Type each solution only​ once.)

We are given the equation:

\[
y^3 + y^2 = 9y + 9
\]

### Step 1: Rearrange the equation
Move all terms to one side:

\[
y^3 + y^2 - 9y - 9 = 0
\]

### Step 2: Factor the equation
We will attempt to factor this cubic equation by grouping.

First, group the terms:

\[
(y^3 + y^2) - (9y + 9) = 0
\]

Factor each group:

\[
y^2(y + 1) - 9(y + 1) = 0
\]

Now, factor out \((y + 1)\):

\[
(y + 1)(y^2 - 9) = 0
\]

### Step 3: Factor further
The second factor is a difference of squares:

\[
y^2 - 9 = (y - 3)(y + 3)
\]

So the equation becomes:

\[
(y + 1)(y - 3)(y + 3) = 0
\]

### Step 4: Solve for \( y \)
Set each factor equal to zero:

1. \( y + 1 = 0 \Rightarrow y = -1 \)
2. \( y - 3 = 0 \Rightarrow y = 3 \)
3. \( y + 3 = 0 \Rightarrow y = -3 \)

### Solution Set:
The solution set is \( \{-3, -1, 3\} \).

These are all real numbers, so no complex numbers are involved.

### Final Answer:
\[
\{-3, -1, 3\}
\]

Would you like further details or explanations?

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Here are 5 related questions for practice:
1. Solve \( y^3 + 4y^2 - 12y = 0 \).
2. Factor the expression \( y^3 + 2y^2 - y - 2 \).
3. Solve the equation \( y^3 - 6y = 0 \).
4. How would you solve a cubic equation using the rational root theorem?
5. What is the solution to \( y^2 - 4y + 4 = 0 \)?

**Tip:** When factoring cubic equations, try grouping terms to find common factors first.

Solve the equation: y^3 + y^2 = 9y + 9 (Simplify your answer. Type an exact answer, using radicals as needed. Express complex numbers in terms of i.)

Discover how to solve the cubic equation y^3 + y^2 = 9y + 9 with a step-by-step guide on factoring and solving for y. This detailed solution includes factoring techniques and the zero product property, perfect for high school students in Grades 9-12.

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