So far we’ve learned two methods for factoring: greatest common factor and difference of squares. Let’s look at another method that will help us when we have perfect cube terms, the sum of cubes method. Whenever you have two perfect cube terms being added (summed), you have a sum of cubes. The first step in factoring using this method is identification:

Identifying a Perfect Cube Term

A perfect cube term can be factored into three identical terms, and nothing more. For example, 8 is a perfect cube because it can be broken down into three identical terms: 2*2*2. You can think of a perfect cube term as any term that’s cube root has no remainder under the radical.

27x³ is a perfect cube because the cube root of 27x³ is 3x.

9b³ is not a perfect cube because the cube root of 9b³ is b cube root of 9, which still contains a root term.

Factoring the Sum of Cubes

Now that you can identify perfect cubes, let’s look at an example:

x³ + y³

As you can see, this is a sum of cubes because x³ and y³ are both perfect cubes and being added (summed) together. Ok, so what can we do with this? We can factor using the following formula:

x³ + y³ = (x+y)(x²-xy+y2)

Every sum of cubes expression will factor using this template. Now let’s use this formula as we look at another example:

8a⁶ + 64b³

Ok, this one looks a little more complicated. First, we need to decide if this is a sum of cubes.

8a⁶ = 2a²*2a²*2a² - Meets criteria for a perfect cube

64b³ = 4b*4b*4b - Meets criteria for a perfect cube

Now that we have identified this expression as a sum of cubes, let’s factor it using the formula we established above:

x³ + y³ = (x+y)(x²-xy+y2)

In our example x³ is now 8a⁶ and y³ is now 64b³, so x=2a² and y=4b

8a⁶ + 64b³ = (2a²+4b)(2a²*2a²-2a²*4b+4b*4b)

Let’s cleanup the expression by multiplying the terms in the second polynomial:

8a⁶ + 64b³ = (2a²+4b)(4a⁴-8a²b+16b²)

And we’re done!