Special Cases - Cubes

Learning Outcomes

Factor special products

Some interesting patterns arise when you are working with cubed quantities within polynomials. Specifically, there are two more special cases to consider:

a^{3}+b^{3}

and

a^{3}-b^{3}

. Let us take a look at how to factor sums and differences of cubes.

Sum of Cubes

The term “cubed” is used to describe a number raised to the third power. In geometry, a cube is a six-sided shape with equal width, length, and height; since all these measures are equal, the volume of a cube with width

x

can be represented by

x^{3}

. (Notice the exponent!) Cubed numbers get large very quickly:

1^{3}=1

2^{3}=8

3^{3}=27

4^{3}=64

, and

5^{3}=125

Before looking at factoring a sum of two cubes, let us look at the possible factors. It turns out that

a^{3}+b^{3}

can actually be factored as

\left(a+b\right)\left(a^{2}–ab+b^{2}\right)

. Check these factors by multiplying.

Example

Does

(a+b)(a^{2}–ab+b^{2})=a^{3}+b^{3}

Answer: Apply the distributive property.

$\left(a\right)\left(a^{2}–ab+b^{2}\right)+\left(b\right)\left(a^{2}–ab+b^{2}\right)$

Multiply by a.

$\left(a^{3}–a^{2}b+ab^{2}\right)+\left(b\right)\left(a^{2}-ab+b^{2}\right)$

Multiply by b.

$\left(a^{3}–a^{2}b+ab^{2}\right)+\left(a^{2}b–ab^{2}+b^{3}\right)$

Rearrange terms in order to combine the like terms.

$a^{3}-a^{2}b+a^{2}b+ab^{2}-ab^{2}+b^{3}$

Simplify.

$a^{3}+b^{3}$

Did you see that? Four of the terms cancelled out, leaving us with the (seemingly) simple binomial

a^{3}+b^{3}

. So, the factors are correct. You can use this pattern to factor binomials in the form

a^{3}+b^{3}

, otherwise known as “the sum of cubes.”

The Sum of Cubes

A binomial in the form

a^{3}+b^{3}

can be factored as

\left(a+b\right)\left(a^{2}–ab+b^{2}\right)

Examples

The factored form of

x^{3}+64

\left(x+4\right)\left(x^{2}–4x+16\right)

. The factored form of

8x^{3}+y^{3}

\left(2x+y\right)\left(4x^{2}–2xy+y^{2}\right)

Example

Factor

x^{3}+8y^{3}

Answer: Identify that this binomial fits the sum of cubes pattern $a^{3}+b^{3}$ . $a=x$ , and $b=2y$ (since $2y\cdot2y\cdot2y=8y^{3}$ ).

$x^{3}+8y^{3}$

Factor the binomial as

\left(a+b\right)\left(a^{2}–ab+b^{2}\right)

, substituting

a=x

and

b=2y

into the expression.

$\left(x+2y\right)\left(x^{2}-x\left(2y\right)+\left(2y\right)^{2}\right)$

Square

(2y)^{2}=4y^{2}

$\left(x+2y\right)\left(x^{2}-x\left(2y\right)+4y^{2}\right)$

Multiply

−x\left(2y\right)=−2xy

(writing the coefficient first).

The factored form is $\left(x+2y\right)\left(x^{2}-2xy+4y^{2}\right)$ .

Now try another one. You should always look for a common factor before you follow any of the patterns for factoring.

Example

Factor

16m^{3}+54n^{3}

Answer: Factor out the common factor $2$ .

$16m^{3}+54n^{3}$

$2\left(8m^{3}+27n^{3}\right)$

8m^{3}

and

27n^{3}

are cubes, so you can factor

8m^{3}+27n^{3}

as the sum of two cubes:

a=2m

and

b=3n

. Factor the binomial

8m^{3}+27n^{3}

substituting

a=2m

and

b=3n

into the expression

\left(a+b\right)\left(a^{2}-ab+b^{2}\right)

$2\left(2m+3n\right)\left[\left(2m\right)^{2}-\left(2m\right)\left(3n\right)+\left(3n\right)^{2}\right]$

Square:

(2m)^{2}=4m^{2}

and

(3n)^{2}=9n^{2}

$2\left(2m+3n\right)\left[4m^{2}-\left(2m\right)\left(3n\right)+9n^{2}\right]$

Multiply

-\left(2m\right)\left(3n\right)=-6mn

The factored form is $2\left(2m+3n\right)\left(4m^{2}-6mn+9n^{2}\right)$ .

Try It

[ohm_question]1952[/ohm_question]

Difference of Cubes

Having seen how binomials in the form

a^{3}+b^{3}

can be factored, it should not come as a surprise that binomials in the form

a^{3}-b^{3}

can be factored in a similar way.

The Difference of Cubes

A binomial in the form

a^{3}–b^{3}

can be factored as

\left(a-b\right)\left(a^{2}+ab+b^{2}\right)

Examples

The factored form of

x^{3}–64

\left(x–4\right)\left(x^{2}+4x+16\right)

. The factored form of

27x^{3}–8y^{3}

\left(3x–2y\left)\right(9x^{2}+6xy+4y^{2}\right)

Notice that the basic construction of the factorization is the same as it is for the sum of cubes; the difference is in the

+

and

–

signs. Take a moment to compare the factored form of

a^{3}+b^{3}

with the factored form of

a^{3}-b^{3}

. Factored form of

a^{3}+b^{3}

\left(a+b\right)\left(a^{2}-ab+b^{2}\right)

Factored form of

a^{3}-b^{3}

\left(a-b\right)\left(a^{2}+ab+b^{2}\right)

This can be tricky to remember because of the different signs. The factored form of

a^{3}+b^{3}

contains a negative, and the factored form of

a^{3}-b^{3}

contains a positive! Some people remember the different forms like this: “Remember one sequence of variables:

a^{3}b^{3}=\left(a\,b\right)\left(a^{2}ab\,b^{2}\right)

. There are

4

missing signs. Whatever the first sign is, it is also the second sign. The third sign is the opposite, and the fourth sign is always

+

.” Try this for yourself. If the first sign is

+

, as in

a^{3}+b^{3}

, according to this strategy, how do you fill in the rest:

\left(a\,b\right)\left(a^{2}ab\,b^{2}\right)

? Does this method help you remember the factored form of

a^{3}+b^{3}

and

a^{3}–b^{3}

? Let us go ahead and look at a couple of examples. Remember to factor out all common factors first.

Example

Factor

8x^{3}–1,000

Answer: Factor out $8$ .

$8(x^{3}–125)$

Identify that the binomial fits the pattern

a^{3}-b^{3}:a=x

, and

b=5

(since

5^{3}=125

). Factor

x^{3}–125

\left(a–b\right)\left(a^{2}+ab+b^{2}\right)

, substituting

a=x

and

b=5

into the expression.

$8\left(x-5\right)\left[x^{2}+\left(x\right)\left(5\right)+5^{2}\right]$

Square the first and last terms, and rewrite

\left(x\right)\left(5\right)

5x

$8\left(x–5\right)\left(x^{2}+5x+25\right)$

Here is one more example. Note that

r^{9}=\left(r^{3}\right)^{3}

and that

8s^{6}=\left(2s^{2}\right)^{3}

Example

Factor

r^{9}-8s^{6}

Answer: Identify this binomial as the difference of two cubes. As shown above, it is. Rewrite $r^{9}$ as $\left(r^{3}\right)^{3}$ and rewrite $8s^{6}$ as $\left(2s^{2}\right)^{3}$ .

$\left(r^{3}\right)^{3}-\left(2s^{2}\right)^{3}$

Now the binomial is written in terms of cubed quantities. Thinking of

a^{3}-b^{3}

a=r^{3}

and

b=2s^{2}

. Factor the binomial as

\left(a-b\right)\left(a^{2}+ab+b^{2}\right)

, substituting

a=r^{3}

and

b=2s^{2}

into the expression.

$\left(r^{3}-2s^{2}\right)\left[\left(r^{3}\right)^{2}+\left(r^{3}\right)\left(2s^{2}\right)+\left(2s^{2}\right)^{2}\right]$

Multiply and square the terms.

$\left(r^{3}-2s^{2}\right)\left(r^{6}+2r^{3}s^{2}+4s^{4}\right)$

In the following two video examples, we show more binomials that can be factored as a sum or difference of cubes. https://youtu.be/tFSEpOB262M https://youtu.be/J_0ctMrl5_0

Try It

[ohm_question]134492[/ohm_question]

You encounter some interesting patterns when factoring. Two special cases—the sum of cubes and the difference of cubes—can help you factor some binomials that have a degree of three (or higher, in some cases). The special cases are:

A binomial in the form $a^{3}+b^{3}$ can be factored as $\left(a+b\right)\left(a^{2}–ab+b^{2}\right)$
A binomial in the form $a^{3}-b^{3}$ can be factored as $\left(a-b\right)\left(a^{2}+ab+b^{2}\right)$

Always remember to factor out any common factors first.

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Special Cases - Cubes

Learning Outcomes

Sum of Cubes

Example

The Sum of Cubes

Examples

Example

Example

Try It

Difference of Cubes

The Difference of Cubes

Examples

Example

Example

Try It

Contribute!

Licenses & Attributions

CC licensed content, Original

CC licensed content, Shared previously

Guias de estudo > ALGEBRA / TRIG I

Special Cases - Cubes

Learning Outcomes

Sum of Cubes

Example

The Sum of Cubes

Examples

Example

Example

Try It

Difference of Cubes

The Difference of Cubes

Examples

Example

Example

Try It

Contribute!

Licenses & Attributions

CC licensed content, Original

CC licensed content, Shared previously