Answer:
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:
[tex](y+3)(y^2-3y+9)=y(y^2-3y+9)+3(y^2-3y+9)[/tex]
[tex]y(y^2-3y+9)+3(y^2-3y+9)=y^3-3y^2+9y+3y^2-9y+27[/tex]
After we add the positive and negative 3y^2 and 9y, they will cancel out and be gone entirely:
[tex]y^3-3y^2+9y+3y^2-9y+27=y^3+27[/tex]
2. You know how you can factor the difference of perfect squares?
As an example:
[tex]a^2-b^2=(a+b)(a-b)[/tex]
Well, not many people know this but you can actually factor both the sum and difference of perfect cubes:
[tex]a^3+b^3=(a+b)(a^2-ab+b^2)[/tex]
[tex]a^3-b^3=(a-b)(a^2+ab+b^2)[/tex]
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
