Special Cases – Cubes | Developmental Math Emporium

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 $(a+b)(a^{2}–ab+b^{2})=a^{3}+b^{3}$ ?

Show Solution

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 $x^{3}+64$ is $\left(x+4\right)\left(x^{2}–4x+16\right)$ .

The factored form of $8x^{3}+y^{3}$ is $\left(2x+y\right)\left(4x^{2}–2xy+y^{2}\right)$ .

Example

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

Show Solution

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

Show Solution

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

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 $x^{3}–64$ is $\left(x–4\right)\left(x^{2}+4x+16\right)$ .

The factored form of $27x^{3}–8y^{3}$ is $\left(3x–2y\left)\right(9x^{2}+6xy+4y^{2}\right)$ .

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 $8x^{3}–1,000$ .

Show Solution

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 $x^{3}–125$ as $\left(a–b\right)\left(a^{2}+ab+b^{2}\right)$ , 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 $\left(x\right)\left(5\right)$ as $5x$ .

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

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

Show Solution

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.

Try It

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.

Module 14: Factoring