Special Cases – Cubes | Intermediate Algebra

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

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.

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 9: Factoring