Factoring Difference of Cubes – Example and Practice Problems

A polynomial of the form a³-b³ is called a difference of cubes. These types of polynomials can be easily factored using a standard pattern. Here, we will learn the process used to factor a difference of cubes.

We will look at several examples with answers to fully master the topic of factoring difference of cubes.

ALGEBRA
factoring difference of cubes

Relevant for

Exploring examples of factoring difference of cubes.

See examples

ALGEBRA
factoring difference of cubes

Relevant for

Exploring examples of factoring difference of cubes.

See examples

How to factor a difference of cubes?

To factor a difference of cubes, we can follow the following steps:

Step 1: Decide if the two terms have a common factor, called the greatest common factor. If so, we factor the greatest common factor from the expression. We must not forget to include the greatest common factor as part of the final answer.

Step 2: Rewrite the original problem as a difference of two perfect cubes.

Step 3: Use the following sentences to write the answer:

a) “Write what you see”

b) “Square – Multiply – Square”

c) “Negative, Positive, Positive”

Step 4: Use all three parts to write the final answer.


Factoring a difference of cubes – Examples with answers

The following difference of cubes factoring examples use the solving process listed above. Each example has a detailed solution that helps to understand the reasoning used to obtain the answer.

EXAMPLE 1

Factor the expression {{x}^3}-8.

Step 1: In this case, the terms do not have any common factors, so we cannot factor initially.

Step 2: We rewrite the original problem as a difference of two perfect cubes:

⇒  {{(x)}^3}-{{(2)}^3}

Step 3: a) If we remove the parentheses and ignore the cubes, we see the expression:

x-2

b) If we square the first term, x, we have {{x}^2}. If we multiply the terms x and 2, we have 2x. If we square the second term, 2, we have 4.

{{x}^2},~~2x,~~4

c) “Negative, Positive, Positive”. These are the signs of the problem.

Step 4: The final answer is:

(x-2)({{x}^2}+2x+4)

EXAMPLE 2

Factor the expression 8{{x}^3}-27.

Step 1: We have no common factors in the terms, so we cannot factor initially.

Step 2: We have to rewrite the original problem as a difference of two perfect cubes:

⇒  {{(2x)}^3}-{{(3)}^3}

Step 3: a) Ignoring the parentheses and cubes, we see the expression:

2x-3

b) Squaring the first term, x, we have 4{{x}^2}. Multiplying the terms 2x and 3, we have 6xSquaring the second term, 3, we have 9.

4{{x}^2},~~6x,~~9

c) “Negative, Positive, Positive”. These are the signs of the problem.

Step 4: The final answer is:

(2x-3)(4{{x}^2}+6x+9)

EXAMPLE 3

Obtain the factorization of 27{{x}^3}-125.

Step 1: This expression also cannot be factored initially since there are no common factors in the terms.

Step 2: We rewrite the original problem as a difference of two perfect cubes:

⇒  {{(3x)}^3}-{{(5)}^3}

Step 3: a) Ignoring the parentheses and cubes, we see the expression:

3x-5

b) Squaring the first term, 3x, we have 9{{x}^2}. By multiplying the terms 3x and 5, we have15xSquaring the second term, 5, we have 25.

9{{x}^2},~~15x,~~25

c) “Negative, Positive, Positive”. These are the signs of the problem.

Step 4: The factored version is:

(3x-5)(9{{x}^2}+15x+25)

EXAMPLE 4

Factor the expression 125{{x}^3} -216{{y}^3}.

Step 1: The terms do not have any common factors, so we continue with the following steps.

Step 2: We write the original expression as a difference of two perfect cubes:

⇒  {{(5x)}^3}-{{(6y)}^3}

Step 3: a) If we ignore the parentheses and cubes, we see the expression:

5x-6y

b) We square the first term, 5x, to get 25{{x}^2}. We multiply the terms 5x and 6y, to get 30xyWe square the second term, 6y, to obtain 36{{y}^2}.

25{{x}^2},~~30xy,~~36{{y}^2}

c) “Negative, Positive, Positive”. These are the signs of the problem.

Step 4: The final answer is:

(5x-6y)(25{{x}^2}+30xy+36{{y}^2})

EXAMPLE 5

Obtain the factorization of the expression 54{{x}^3}-16{{y}^3}.

Step 1: This expression does have common factors. We can extract the 2 from both terms:

2(27{{x}^3}-8{{y}^3})

Step 2: We rewrite the expression obtained as a difference of two perfect cubes:

⇒  2({{(3x)}^3}-{{(2y)}^3})

Step 3: a) By removing the parentheses and ignoring the cubes, we have:

3x-2y

b) If we square the first term, 3x, we have 9{{x}^2}. If we multiply the terms 3x and 2y, we have 6xyIf we square the second term, 2y, we have 4{{y}^2}.

9{{x}^2},~~6xy,~~4{{y}^2}

c) “Negative, Positive, Positive”. These are the signs of the problem.

Step 4: We obtained the following factorization:

2(3x-2y)(9{{x}^2}+6xy+4{{y}^2})

EXAMPLE 6

Factor the expression 1-125{{x}^3}{{y}^3}.

Step 1: We have no common factors.

Step 2: By rewriting the expression as a difference of two perfect cubes, we have:

⇒  {{(1)}^3}-{{(5xy)}^3}

Step 3: a) If we remove the parentheses and ignore the cubes, we see the expression:

1-5xy

b) Squaring the first term, 1, we have 1. Multiplying the terms 1 and 5xy, we get 5xySquaring the second term, 5xy, we have 25{{x}^2}{{y}^2}.

1,~~5xy,~~25{{x}^2}{{y}^2}

c) “Negative, Positive, Positive”. These are the signs of the problem.

Step 4: The final answer is:

(1-5xy)(1+5xy+25{{x}^2}{{y}^2})


Factoring difference of cubes – Practice problems

Test what you have learned about factoring the difference of cubes with the following problems. Solve the exercises and choose an answer. Check the selected answer to see if it is correct.

Factor the expression {{x}^3}-125.

Choose an answer






Factor the expression 127{{x}^3}-27.

Choose an answer






Factor the expression 64{{x}^3}-343{{y}^3}.

Choose an answer






Factor the expression 2{{x}^3}-128{{y}^3}.

Choose an answer








See also

Interested in learning more about the factorization of polynomials? Take a look at these pages:

Learn mathematics with our additional resources in different topics

LEARN MORE