Factoring Binomials Cubed – Examples and Practice Problems

Cubed binomials can be in the form of a sum of cubes or a difference of cubes. These binomials can be easily factored using general formulas. The process used to factor both binomials is similar with a simple change in the signs of the final expression.

Here, we will review the process used to obtain the factorization of binomials cubed. In addition, we will look at several practice problems to understand the application of this process.

ALGEBRA
factoring difference of cubes

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Learning about factoring binomials cubed with examples.

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ALGEBRA
factoring difference of cubes

Relevant for

Learning about factoring binomials cubed with examples.

See examples

How to factor binomials cubed?

We can have two types of binomials cubed, a difference or a sum. The sum of cubes is an expression with the general form {{a}^3}+{{b}^3} and a difference of cubes is an expression with the general form {{a}^3}-{{b}^3}.

To factor binomials cubed, we can follow the following steps:

Step 1: Factor the common factor of the terms if it exists to obtain a simpler expression. We must not forget to include the common factor in the final answer.

Step 2: We have to rewrite the expression as a sum or difference of two perfect cubes.

Step 3: We can write the answer using the following sentences:

a) “Write what you see”

b) “Square – Multiply – Square”

c) If it is a sum of cubes, we have the signs “Positive, Negative, Positive” and if it is a difference of cubes, we have the signs “Negative, Positive, Positive”

Step 4: We join the resulting parts to obtain the final factored expression.


Factoring binomials cubed – Examples with answers

The following examples of factoring binomials cubed apply the solving process detailed above. It is recommended that you try to solve the exercises yourself before looking at the solution.

EXAMPLE 1

Factor the binomial {{x}^3}+8.

Step 1: We don’t have any common factors to factor, so we can’t simplify.

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

⇒  {{(x)}^3}+{{(2)}^3}

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

x+2

b) Squaring the first term, x, we get {{x}^2}. By multiplying the terms x and 2, we get 2xSquaring the second term, 2, we have 4.

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

c) This is a sum of binomials cubed, so the signs are “Positive, Negative, Positive”.

Step 4: The factored expression is:

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

EXAMPLE 2

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

Step 1: Here, we don’t have common factors either, so we can’t factor initially.

Step 2: Now, we write the original problem as a difference of two perfect cubes:

⇒  {{(x)}^3}-{{(3)}^3}

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

x-3

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

{{x}^2},~~3x,~~9

c) This is a difference of cubes, so the signs we have to use are “Negative, Positive, Positive”.

Step 4: The factored expression is:

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

EXAMPLE 3

Obtain the factorization of the sum of cubes 8{{x}^3}+125.

Step 1: We cannot factor this expression initially as there are no common factors.

Step 2: We can rewrite the expression as a sum of two perfect cubes:

⇒  {{(2x)}^3}-{{(5)}^3}

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

2x-5

b) If we square the first term, 2x, we have 4{{x}^2}. If we multiply the terms 2x and 5, we have 10xIf we square the second term, 5, we have 25.

4{{x}^2},~~10x,~~25

c) For a sum of cubes, we must use the signs “Positive, Negative, Positive”.

Step 4: The factored expression is:

(2x+5)(4{{x}^2}-10x+25)

EXAMPLE 4

Factor the difference of cubes 27{{x}^3}-216{{y}^3}.

Step 1: We cannot factor initially because we do not have common factors.

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

⇒  {{(3x)}^3}-{{(6y)}^3}

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

3x-6y

b) Squaring the first term, 3x, we have 9{{x}^2}. By multiplying the terms 3x and 6y, we get 18xy. If we square the second term, 6y, we get 36{{y}^2}.

9{{x}^2},~~18xy,~~36{{y}^2}

c) For a difference of cubes, we have the signs “Negative, Positive, Positive”.

Step 4: The final factored expression is:

(3x-6y)(9{{x}^2}+18xy+36{{y}^2})

EXAMPLE 5

Facor the sum of cubes 54{{x}^3}+16{{y}^3}.

Step 1: Here, we do have a common factor, 2. We can extract the 2 from both terms:

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

Step 2: Now, we can write the expression as a sum of two perfect cubes:

⇒  2({{(3x)}^3}+{{(2y)}^3})

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

3x+2y

b) The first term, 3x, when squared is equal to 9{{x}^2}. The product of the terms 3x and 2is 6xyThe square of the second term, 2y, is 4{{y}^2}.

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

c) For a sum of cubes, we have the signs “Positive, Negative, Positive”.

Step 4: We obtained the following factorization:

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

EXAMPLE 6

Obtain the factorization of 8-27{{x}^3}{{y}^3}.

Step 1: We have no common factor in the terms.

Step 2: If we rewrite the expression as a difference of two perfect cubes, we have:

⇒  {{(2)}^3}-{{(3xy)}^3}

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

2-3xy

b) The first term, 2, when squared is 4. The product of the terms 2 and 3xy is 6xyThe second term, 3xy squared is 9{{x}^2}{{y}^2}.

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

c) For a difference of cubes, the signs are “Negative, Positive, Positive”.

Step 4: The factored expression is:

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


Factoring binomials cubed – Practice problems

Practice what you have learned about factoring binomials cubed with the following problems. If you need help, you can look at the solved examples shown above.

Factor the sum of cubes {{x}^3}+125.

Choose an answer






Factor the difference of cubes 125{{x}^3}-27.

Choose an answer






Factor the sum of cubes 64{{x}^3}+343{{y}^3}.

Choose an answer






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

Choose an answer







See also

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