Special Cases: Cubes

 

Learning Objective(s)

·         Factor the sum of cubes.

·         Factor the difference of cubes.

 

 

In many ways, factoring is about patterns—if you recognize the patterns that numbers make when they are multiplied together, you can use those patterns to separate these numbers into their individual factors.

 

Some interesting patterns arise when you are working with cubed quantities within polynomials. Specifically, there are two more special cases to consider: a³ + b³ and a³ – b³.

 

Let’s take a look at how to factor sums and differences 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 x³. (Notice the exponent!)

 

Cubed numbers get large very quickly. 1³ = 1, 2³ = 8, 3³ = 27, 4³ = 64, and 5³ = 125.

 

Before looking at factoring a sum of two cubes, let’s look at the possible factors.

 

It turns out that a³ + b³ can actually be factored as (a + b)(a² – ab + b²). Let’s check these factors by multiplying.

 

 

Did you see that? Four of the terms cancelled out, leaving us with the (seemingly) simple binomial a³ + b³. So, the factors are correct.

 

You can use this pattern to factor binomials in the form a³ + b³, otherwise known as “the sum of cubes.”

 

The Sum of Cubes   A binomial in the form a3 + b3 can be factored as (a + b)(a2 – ab + b2).   Examples: The factored form of x3 + 64 is (x + 4)(x2 – 4x + 16). The factored form of 8x3 + y3 is (2x + y)(4x2 – 2xy + y2).

 

 

 

 

And that’s it. The binomial x³ + 8y³ can be factored as (x + 2y)(x² – 2xy + 4y²)! Let’s try another one.

 

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

 

 

 

 

Factor 125x3 + 64.   A) (5x + 64)(25x2 – 125x + 16)   B) (5x + 4)(25x2 – 20x + 16)   C) (x + 4)(x2 – 2x + 16)   D) (5x + 4)(25x2 + 20x – 64)   Show/Hide Answer A) (5x + 64)(25x2 – 125x + 16) Incorrect. Check your values for a and b here. b3 = 64, so what is b? The correct answer is (5x + 4)(25x2 – 20x + 16).   B) (5x + 4)(25x2 – 20x + 16) Correct. 5x is the cube root of 125x3, and 4 is the cube root of 64. Substituting these values for a and b, you find (5x + 4)(25x2 – 20x + 16).   C) (x + 4)(x2 – 2x + 16) Incorrect. Check your values for a and b here. a3 = 125x3, so what is a? The correct answer is (5x + 4)(25x2 – 20x + 16).   D) (5x + 4)(25x2 + 20x – 64) Incorrect. Check the mathematical signs; the b2 term is positive, not negative, when factoring a sum of cubes. The correct answer is (5x + 4)(25x2 – 20x + 16).

 

 

 

Having seen how binomials in the form a³ + b³ can be factored, it should not come as a surprise that binomials in the form a³ – b³ can be factored in a similar way.

 

The Difference of Cubes   A binomial in the form a3 – b3 can be factored as (a – b)(a2 + ab + b2).   Examples: The factored form of x3 – 64 is (x – 4)(x2 + 4x + 16). The factored form of 27x3 – 8y3 is (3x – 2y)(9x2 + 6xy + 4y2).

 

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 a³ + b³ with the factored form of a³ – b³.

 

 

This can be tricky to remember because of the different signs—the factored form of a³ + b³ contains a negative, and the factored form of a³ – b³ contains a positive! Some people remember the different forms like this:

 

“Remember one sequence of variables: a³   b³  =  (a   b)(a²   ab   b²). 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 a³ + b³, according to this strategy how do you fill in the rest: (a   b)(a²   ab   b²)? Does this method help you remember the factored form of a³ + b³ and a³ – b³?

 

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

 

 

 

 

Let’s see what happens if you don’t factor out the common factor first. In this example, it can still be factored as the difference of two cubes. However, the factored form still has common factors, which need to be factored out.

 

 

 

 

As you can see, this last example still worked, but required a couple of extra steps. It is always a good idea to factor out all common factors first. In some cases, the only efficient way to factor the binomial is to factor out the common factors first.

 

Here is one more example. Note that r⁹ =  (r³)³ and that  8s⁶ = (2s²)³.

 

 

 

 

Using the difference of cubes, identify the product of 3(x – 3y)(x2 + 3xy + 9y2).   A) x3 – y3   B) 3x – 81y   C) 3x3 + 81y3   D) 3x3 – 81y3   Show/Hide Answer A) x3 – y3 Incorrect. If this were true, the expression shown above would be (x – y)(x2 + xy + y2). The correct answer is 3x3 – 81y3.   B) 3x – 81y Incorrect. Neither of the terms in this binomial is a cubed number! The correct answer is 3x3 – 81y3.   C) 3x3 + 81y3 Incorrect. Check your signs. If this expression falls into the difference of cubes category, the symbol between 3x3 and 81y3 should be –. The correct answer is 3x3 – 81y3.   D) 3x3 – 81y3 Correct. Recognizing that this expression is in the form (a – b)(a2 + ab + b2), you find a = x and b = 3y. This means that the resulting a3 – b3 monomial is x3 – 27y3. It also needs to be multiplied by the coefficient 3: 3x3 – 81y3.

 

 

 

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 a³ + b³ can be factored as (a + b)(a² – ab + b²)
• A binomial in the form a³ – b³ can be factored as (a – b)(a² + ab + b²)

 

Always remember to factor out any common factors first.