Question

(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

Answer 1

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:

$(y+3)(y^2-3y+9)=y(y^2-3y+9)+3(y^2-3y+9)$

$y(y^2-3y+9)+3(y^2-3y+9)=y^3-3y^2+9y+3y^2-9y+27$

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

$y^3-3y^2+9y+3y^2-9y+27=y^3+27$

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

As an example:

$a^2-b^2=(a+b)(a-b)$

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

$a^3+b^3=(a+b)(a^2-ab+b^2)$

$a^3-b^3=(a-b)(a^2+ab+b^2)$

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