(Y+3)(y^2-3y+9)

A. Y^3+27
B. Y^3-27
C. Y^3-6y^2+27
D. Y^3+6y^2+27

A) y^3+27

Step-by-step explanation:

There are two ways of solving this problem:

1. Recognizing this as the factored form of the sum of perfect cubes

2. Distribute and add the like terms.

1. In order to distribute we must multiply y by y^2-3y+9, and then 3 by y^2-3y+9:

After we add the positive and negative 3y^2 and 9y, they will cancel out and be gone entirely:

2. You know how you can factor the difference of perfect squares?

As an example:

Well, not many people know this but you can actually factor both the sum and difference of perfect cubes:

Because we have these identities, we can easily establish here that we have the sum of perfect cubes, and that (y+3)(y^2-3y+9)= y^3+3^3 = y^3+27

60

step-by-step explanation:

2

step-by-step explanation: