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High School Algebra I: Help and Review25 chapters | 292 lessons

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Lesson Transcript

Instructor:
*Miriam Snare*

Miriam has taught middle- and high-school math for over 10 years and has a master's degree in Curriculum and Instruction.

In this lesson, you will learn how to factor sums and differences of perfect cube binomials. You will learn to use the formula that gives you the factors through a few examples.

A **perfect cube** is any expression that can represent the volume of a cube. Imagine reaching into a toy bin and pulling out several building blocks and then try to arrange the blocks into a cube. Let's say you pulled out 4 blocks. You can make a square of 2 blocks by 2 blocks with 4 blocks, but it will only have a height of 1 block, so that does not make a cube. In order to create a cube, you would need to pull out 4 more blocks to layer on top of the square you made, like this:

Now, you have a volume of 8 blocks that make a cube whose length, width, and height are all 2. So, 8 is a perfect cube number because 2 x 2 x 2 = 8 or 2^3 = 8.

Say you wanted to increase the size of the cube to be 3 blocks along each side, to look like this:

Then, you would need 19 more blocks for a total of 27, because 3 x 3 x 3 = 3^3 = 27.

All of the perfect cubes result from raising an **integer**, a whole number or its negative counterpart, or a variable to the third power. These expressions are called cubes because the volume of a cube is found by raising the length of its side to the third power.

A **perfect cube binomial** is an expression consisting of two terms that are perfect cubes. These terms are either added or subtracted from each other.

**Factors** are expressions that are multiplied together. So, **factoring** asks you to break down an expression into more simple expressions that are multiplied. The factors of a perfect cube binomial may not look very simple because they end up being a **binomial**, two terms added or subtracted, times a **trinomial**, three terms added or subtracted. However, there is a formula that gives you the factors quite easily.

When faced with assorted factoring problems, you have different methods to tackle different expressions. Make sure that you are applying the correct strategy for each type of problem. A perfect cube binomial is fairly easy to identify, and it can always be factored using this formula. Your first clues will be that there are always two terms and the variables have exponents that are multiples of 3. Once you have spotted those characteristics, you want to check that the **coefficient** (number multiplied to variables) and **constant** (number with no variables) of both terms are perfect cube numbers.

Once you are sure that you are factoring a perfect cube binomial, you are going to use this formula:

Here's a hint: If you have a difference of cubes in the form *a*^3 - *b*^3, just think of it as *a*^3 + (-*b*^3). That way you only have to memorize this one formula.

Let's take a look at some examples:

Factor: *x*^3 + 64

In this problem, we see that two terms have been added together, making it a binomial. Both of the terms are perfect cubes. We can tell that from the power of 3 on the *x*, and we know 64 is a perfect cube because a cube measuring 4 x 4 x 4 has a volume of 4^3 = 64. To factor the expression, we do the following:

- We determine what the
*a*and*b*for our formula are. The first term is easy,*x*is the*a*. In the second term, 64 = 4^3, so*b*is 4. - Then, we substitute those values into the formula: (
*x*+ 4)(*x*^2 - (*x*)(4) + 4^2). - We can rewrite the (
*x*)(4) as 4*x*, and evaluate the exponent in the last term to clean it up. Our final factored form is: (*x*+ 4)(*x*^2 - 4*x*+ 16).

Here's another example. Factor: 8*x*^3 - 125

In this problem, we see that two terms subtracted from each other, making it a binomial. The power of 3 on the *x* tells us that we are most likely dealing with a perfect cube binomial. Let's check the numbers to be sure they are perfect cubes. The number 8 is a perfect cube because, as we discussed at the beginning of this lesson, 2^3 = 8. The number 125 is a perfect cube because a cube measuring 5 x 5 x 5 would have a volume of 5^3 = 125.

To factor the expression, we do the following:

- We determine what the
*a*and*b*for our formula are. This time in the first term, 8*x*^3 = (2*x*)^3. Therefore,*a*is 2*x*. In the second term is negative this time, -125 = (-5)^3, so*b*is -5. - Then, we substitute those values into the formula: (2
*x*+ (-5))((2*x*)^2 - (2*x*)(-5) + (-5)^2). - We need to clean up the terms a bit. In the binomial factor (2
*x*+ (-5)), we can combine the plus sign and the negative to write it as (2*x*- 5). In the trinomial factor, the first term (2*x*)^2 can be evaluated to be 4*x*^2. In the next term, we can multiply the (2*x*) and the (-5) together to get (-10*x*). Since that term was subtracted, instead of using a double-negative -(-10*x*), we can write +10*x*. In the final term, we are squaring (-5), which gives us 25. So, the final factored form should be (2*x*- 5)(4*x*^2 + 10*x*+ 25).

Here's a final example. Factor: 27*x*^6 - 343*y*^3

In this problem, we have two terms subtracted from each other, making it a binomial. Let's check that everything is a perfect cube. 27 is a perfect cube because, as we discussed at the beginning of this lesson, 3^3 = 27. 343 might take a little figuring out, but it is a perfect cube because a cube measuring 7 x 7 x 7 would have a volume of 7^3 = 343. Now, let's look at the exponents. The *y*^3 is definitely a perfect cube, but what about the *x*^6? Remember that any power that is a multiple of 3 is also a perfect cube. In this case, we need to think of *x*^6 as (*x*^2)^3.

To factor the expression, we do the following:

- We determine what the
*a*and*b*for our formula are. In the first term, 27*x*^6 = (3*x*^2)^3. Therefore,*a*is 3*x*^2. The second term is negative this time, -343*y*^3 = (-7*y*)^3, so*b*is -7*y.* - Then, we substitute those values into the formula: (3
*x*^2 + (-7*y*))(( 3*x*^2)^2 - (3*x*^2)( -7*y*) + (-7*y*)^2. - Just as in the previous example, we have to clean it up the terms. In the binomial factor, we can combine the plus and the negative to write (3
*x*^2 - 7*y*). In the trinomial factor, the first term (3*x*^2)^2 can be evaluated to be 9*x*^4. In the next term, we can multiply the (3*x*^2) and the (-7*y*) together to get (-21(*x*^2)*y*). Since that term was subtracted, instead of using a double-negative, we can write +21(*x*^2)*y*. In the final term, we are squaring (-7*y*), which gives us 49*y*^2. So, the final factored form should be (3*x*^2 -7*y*)(9*x*^4 + 21(*x*^2)*y*+ 49*y*^2).

A **perfect cube binomial** has two terms that are perfect cubes. To factor one of these binomials, you first identify the *a* and *b* for the formula by figuring out what was cubed in each term. Then, you fill in the formula:

*a*^3 + *b*^3 = (*a* + *b*)(*a*^2 - *ab* + *b*^2)

Finally, you clean up any multiplications and exponents. Your answer will always have two factors, one binomial and one trinomial.

- perfect cube: any expression that can represent the volume of a cube
- integer: a whole number or its negative counterpart
- perfect cube binomial: an expression consisting of two terms that are perfect cubes
- factors: expressions that are multiplied together
- factoring: breaking down an expression into more simple expressions that are multiplied
- binomial: two terms added or subtracted
- trinomial: three terms added or subtracted
- coefficient: a number multiplied to variables
- constant: a number with no variables

After viewing this lesson, you should be able to meet the following goals:

- Define a perfect cube binomial
- Factor a perfect cube binomial

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High School Algebra I: Help and Review25 chapters | 292 lessons

- What is Factoring in Algebra? - Definition & Example 5:32
- How to Find the Prime Factorization of a Number 5:36
- Using Prime Factorizations to Find the Least Common Multiples 7:28
- Equivalent Expressions and Fraction Notation 5:46
- Using Fraction Notation: Addition, Subtraction, Multiplication & Division 6:12
- Factoring Out Variables: Instructions & Examples 6:46
- Combining Numbers and Variables When Factoring 6:35
- Transforming Factoring Into A Division Problem 5:11
- Factoring By Grouping: Steps, Verification & Examples 7:46
- How to Factor a Perfect Cube: Formula & Examples 9:28
- Go to High School Algebra - Factoring: Help and Review

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