# Difference of Cubes

Reference > Mathematics > Algebra > Factoring Higher Degree PolynomialsPreviously we looked at a type of factoring problem which involved two perfect squares. On this page, we'll learn a method for factoring binomials in which the first term is a perfect cube, the second term is a perfect cube, and the two terms are being subtracted.

Before we get started with the factoring rule, let's do a little bit of multiplication practice. Can you multiply the polynomials below?

(a - b)(a^{2} +ab + b^{2})

a(a^{2} +ab + a^{2}) - b(a^{2} +ab + b^{2})

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

a^{3} - b^{3}

Interesting; if we multiply those two polynomials, we end up with a perfect cube subtracted from another perfect cube, which is exactly what we're looking for. Let's formalize this into a factoring rule:

**Difference of Cubes**

a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})

You should jot this down somewhere, and memorize it, because you'll use this any time you see this sort of expression to factor.

**Example One**

Factor 8x^{3} - 27

**Solution**

As always, we begin by checking to see if there is a common factor which can be factored out. Since there is not, we ask ourselves, "Are the two terms perfect cubes?" Yes they are! The first term is the cube of 2x, and the second is the cube of 3 (we ignore the negative sign for now, because our rule takes care of that for us!).

So in our rule, a = 2x, and b = 3

8x^{3} - 27 = (2x - 3)((2x)^{2} + 2x(3) + 3^{2})

8x^{3} - 27 = (2x - 3)(4x^{2} + 6x + 9)

**Example Two**

Factor 16x^{4} - 2x

**Solution**

First we note that we can factor out 2x:

16x^{4} - 2x = 2x(8x^{3} - 1)

Now we note that both 8x^{3} and 1 are pefect cubes, so a = 2x and b = 1.

16x^{4} - 2x = 2x(2x - 1)((2x)^{2} + 2x(1) + 1^{2})

16x^{4} - 2x = 2x(2x - 1)(4x^{2} + 2x + 1)

**Example Three**

Factor 27x^{6} - 1000y^{9}

**Solution**

There are no common factors to bring out, so we begin by noting that these are perfect cubes, and a = 3x^{2} and b = 10y^{3}

27x^{6} - 1000y^{9} = (3x^{2} - 10y^{3})((3x^{2})^{2} +3x^{2}(10y^{3}) + (10y^{3})^{2})

27x^{6} - 1000y^{9} = (3x^{2} - 10y^{3})(9x^{4} + 30x^{2}y^{3} + 100y^{6})

**Checking Your Work**

If you want to check your work, you can (of course!) multiply your entire answer out, to see if it gets you back to the original problem. If you don't have time to do that, there is a quick "check" method which - while it doesn't guarantee that you have the right answer - will help to check the "problem spots" that students often have. In your answer, multiply the first term of each of your two factors, and multiply the last term of each of your two factors.

In the previous example, that would be:

3x^{2}(9x^{4}) = 27x^{6}

10y^{3}(100y^{6}) = 1000y^{9}

Notice that those two quantities match the two terms in your original binomial. If they don't match, you've done something wrong, and you need to go back to the drawing board!

## Questions

^{3}- 64

^{3}- 81y

^{3}

^{5}- 8x

^{2}y

^{3}

^{3r}- y

^{6q}

^{3}b

^{6}