How do you factor these?

Question

1. 8y^3+27
2. 27m^3-8

Answer 1

1. 8y^3 + 27 > Use the sum of cubes formula which is...

(a+b)(a^2 - ab + b^2)

First: express the terms in lowest terms...

(2y)(2y)(2y) + (3)(3)(3)

Sec: you have two variables...

a = 2y
b = 3

Third: replace the values with the corresponding variables...

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

(2y + 3)((2y)(2y) - (2y)(3) + 3*3)

(2y + 3)(4y^2 - 6y + 9)


2. 27m^3 - 8 > Use the differences of cubes formula which is...

(a - b)(a^2 + ab + b^2)

First: express the terms in lowest terms...

(3m)(3m)(3m) - (2)(2)(2)

Sec: you have two variables...

a = 3m
b = 2

Third: replace the values with the corresponding variables...

(3m - 2)((3m)^2 + (3m)(2) + 2^2)

(3m - 2)((3m)(3m) + (3m)(2) + 2*2)

(3m - 2)(9m^2 + 6m + 4)

Answer 2

These are both known as sums and differences of cubes, because
8y^3 is the same as (2y)^3, and
27 is the same as 3^3.

To factor sums and differences of cubes, you do the following:
1) Take the cube root of each term, and place in one bracket.
2) In the second bracket, perform the following:
a) Square the first
b) Negative product
c) Square the last.

Let's follow these instructions.

1) 8y^3 + 27

Taking the cube root of each term,

(2y + 3) (? + ? + ?)

"Square the first"

We take the square of the first value in the first brackets, and place in the second brackets.

(2y + 3) (4y^2 + ? + ?)

"Negative product."
Multiply 2y and 3 together, and then take the negative.
(2y)(3) = 6y, negated is -6y.

(2y + 3) (4y^2 - 6y + ?)

"Square the last."
Square the second term in the first set of brackets.

(2y + 3)(4y^2 - 6y + 9)

Now, do the same for the next question.

Answer 3

1. (2y +3) (4y^2 - 6y +9)

2. (3m-2)(9m^2+6m+4)