# 6.3: Factor Sums and Differences of Cubes - Mathematics

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If we check to see whether either term is a cube,

we can see that both terms are perfect cubes. The difference of cubes formula says . a^3-b^3. is always factored as

Since in this case . a=c. and . b=2b^4. we get

We know we’re dealing with the difference of cubes, because we have two perfect cubes separated by subtraction.

If we check to see whether either term is a cube,

we can see that both terms are perfect cubes. The difference of cubes formula says . a^3-b^3. is always factored as

The variable . a. will be the cube root of the first term, and the variable . b. will be the cube root of the second term. So

We can check our work by distributing each term in the binomial factor over each term in the trinomial factor.

## 6.3 Factor Special Products

We have seen that some binomials and trinomials result from special products—squaring binomials and multiplying conjugates. If you learn to recognize these kinds of polynomials, you can use the special products patterns to factor them much more quickly.

### Factor Perfect Square Trinomials

Some trinomials are perfect squares. They result from multiplying a binomial times itself. We squared a binomial using the Binomial Squares pattern in a previous chapter.

In this chapter, you will start with a perfect square trinomial and factor it into its prime factors.

Here is the pattern—the reverse of the binomial squares pattern.

### Perfect Square Trinomials Pattern

If a and b are real numbers

To make use of this pattern, you have to recognize that a given trinomial fits it. Check first to see if the leading coefficient is a perfect square, a 2 . a 2 . Next check that the last term is a perfect square, b 2 . b 2 . Then check the middle term—is it the product, 2 a b ? 2 a b ? If everything checks, you can easily write the factors.

### Example 6.23

#### Solution

The sign of the middle term determines which pattern we will use. When the middle term is negative, we use the pattern a 2 − 2 a b + b 2 , a 2 − 2 a b + b 2 , which factors to ( a − b ) 2 . ( a − b ) 2 .

The steps are summarized here.

### How To

#### Factor perfect square trinomials.

Step 1. Does the trinomial fit the pattern? a 2 + 2 a b + b 2 a 2 − 2 a b + b 2 Is the first term a perfect square? ( a ) 2 ( a ) 2 Write it as a square. Is the last term a perfect square? ( a ) 2 ( b ) 2 ( a ) 2 ( b ) 2 Write it as a square. Check the middle term. Is it 2 a b ? ( a ) 2 ↘ 2 · a · b ↙ ( b ) 2 ( a ) 2 ↘ 2 · a · b ↙ ( b ) 2 Step 2. Write the square of the binomial. ( a + b ) 2 ( a − b ) 2 Step 3. Check by multiplying. Step 1. Does the trinomial fit the pattern? a 2 + 2 a b + b 2 a 2 − 2 a b + b 2 Is the first term a perfect square? ( a ) 2 ( a ) 2 Write it as a square. Is the last term a perfect square? ( a ) 2 ( b ) 2 ( a ) 2 ( b ) 2 Write it as a square. Check the middle term. Is it 2 a b ? ( a ) 2 ↘ 2 · a · b ↙ ( b ) 2 ( a ) 2 ↘ 2 · a · b ↙ ( b ) 2 Step 2. Write the square of the binomial. ( a + b ) 2 ( a − b ) 2 Step 3. Check by multiplying.

We’ll work one now where the middle term is negative.

### Example 6.24

#### Solution

The first and last terms are squares. See if the middle term fits the pattern of a perfect square trinomial. The middle term is negative, so the binomial square would be ( a − b ) 2 . ( a − b ) 2 .

The next example will be a perfect square trinomial with two variables.

### Example 6.25

Factor: 36 x 2 + 84 x y + 49 y 2 . 36 x 2 + 84 x y + 49 y 2 .

#### Solution

 Test each term to verify the pattern. Factor. Check by multiplying. ( 6 x + 7 y ) 2 ( 6 x ) 2 + 2 · 6 x · 7 y + ( 7 y ) 2 36 x 2 + 84 x y + 49 y 2 ✓ ( 6 x + 7 y ) 2 ( 6 x ) 2 + 2 · 6 x · 7 y + ( 7 y ) 2 36 x 2 + 84 x y + 49 y 2 ✓

Factor: 49 x 2 + 84 x y + 36 y 2 . 49 x 2 + 84 x y + 36 y 2 .

Factor: 64 m 2 + 112 m n + 49 n 2 . 64 m 2 + 112 m n + 49 n 2 .

Remember the first step in factoring is to look for a greatest common factor. Perfect square trinomials may have a GCF in all three terms and it should be factored out first. And, sometimes, once the GCF has been factored, you will recognize a perfect square trinomial.

### Example 6.26

Factor: 100 x 2 y − 80 x y + 16 y . 100 x 2 y − 80 x y + 16 y .

#### Solution

Remember: Keep the factor 4y in the final product.

Check:

4 y ( 5 x − 2 ) 2 4 y [ ( 5 x ) 2 − 2 · 5 x · 2 + 2 2 ] 4 y ( 25 x 2 − 20 x + 4 ) 100 x 2 y − 80 x y + 16 y ✓ 4 y ( 5 x − 2 ) 2 4 y [ ( 5 x ) 2 − 2 · 5 x · 2 + 2 2 ] 4 y ( 25 x 2 − 20 x + 4 ) 100 x 2 y − 80 x y + 16 y ✓

Factor: 8 x 2 y − 24 x y + 18 y . 8 x 2 y − 24 x y + 18 y .

Factor: 27 p 2 q + 90 p q + 75 q . 27 p 2 q + 90 p q + 75 q .

### Factor Differences of Squares

The other special product you saw in the previous chapter was the Product of Conjugates pattern. You used this to multiply two binomials that were conjugates. Here’s an example:

A difference of squares factors to a product of conjugates.

### Difference of Squares Pattern

If a and b are real numbers,

Remember, “difference” refers to subtraction. So, to use this pattern you must make sure you have a binomial in which two squares are being subtracted.

### How To

#### Factor differences of squares.

Step 1. Does the binomial fit the pattern? a 2 − b 2 Is this a difference? ____ − ____ Are the first and last terms perfect squares? Step 2. Write them as squares. ( a ) 2 − ( b ) 2 Step 3. Write the product of conjugates. ( a − b ) ( a + b ) Step 4. Check by multiplying. Step 1. Does the binomial fit the pattern? a 2 − b 2 Is this a difference? ____ − ____ Are the first and last terms perfect squares? Step 2. Write them as squares. ( a ) 2 − ( b ) 2 Step 3. Write the product of conjugates. ( a − b ) ( a + b ) Step 4. Check by multiplying.

It is important to remember that sums of squares do not factor into a product of binomials. There are no binomial factors that multiply together to get a sum of squares. After removing any GCF, the expression a 2 + b 2 a 2 + b 2 is prime!

The next example shows variables in both terms.

### Example 6.28

#### Solution

As always, you should look for a common factor first whenever you have an expression to factor. Sometimes a common factor may “disguise” the difference of squares and you won’t recognize the perfect squares until you factor the GCF.

Also, to completely factor the binomial in the next example, we’ll factor a difference of squares twice!

### Example 6.29

Factor: 48 x 4 y 2 − 243 y 2 . 48 x 4 y 2 − 243 y 2 .

#### Solution

Factor: 7 a 4 c 2 − 7 b 4 c 2 . 7 a 4 c 2 − 7 b 4 c 2 .

The next example has a polynomial with 4 terms. So far, when this occurred we grouped the terms in twos and factored from there. Here we will notice that the first three terms form a perfect square trinomial.

### Example 6.30

#### Solution

Notice that the first three terms form a perfect square trinomial.

 Factor by grouping the first three terms. Use the perfect square trinomial pattern. Is this a difference of squares? Yes. Yes—write them as squares. Factor as the product of conjugates.

You may want to rewrite the solution as ( x − y − 3 ) ( x + y − 3 ) . ( x − y − 3 ) ( x + y − 3 ) .

Factor: x 2 − 10 x + 25 − y 2 . x 2 − 10 x + 25 − y 2 .

Factor: x 2 + 6 x + 9 − 4 y 2 . x 2 + 6 x + 9 − 4 y 2 .

### Factor Sums and Differences of Cubes

There is another special pattern for factoring, one that we did not use when we multiplied polynomials. This is the pattern for the sum and difference of cubes. We will write these formulas first and then check them by multiplication.

We’ll check the first pattern and leave the second to you.

### Sum and Difference of Cubes Pattern

The two patterns look very similar, don’t they? But notice the signs in the factors. The sign of the binomial factor matches the sign in the original binomial. And the sign of the middle term of the trinomial factor is the opposite of the sign in the original binomial. If you recognize the pattern of the signs, it may help you memorize the patterns.

The trinomial factor in the sum and difference of cubes pattern cannot be factored.

It will be very helpful if you learn to recognize the cubes of the integers from 1 to 10, just like you have learned to recognize squares. We have listed the cubes of the integers from 1 to 10 in Table 6.1.

### How To

#### Factor the sum or difference of cubes.

1. Step 1. Does the binomial fit the sum or difference of cubes pattern?
Is it a sum or difference?
Are the first and last terms perfect cubes?
2. Step 2. Write them as cubes.
3. Step 3. Use either the sum or difference of cubes pattern.
4. Step 4. Simplify inside the parentheses.
5. Step 5. Check by multiplying the factors.

### Example 6.32

#### Solution

 This binomial is a difference. The first and last terms are perfect cubes. Write the terms as cubes. Use the difference of cubes pattern. Simplify. Check by multiplying. We’ll leave the check to you.

In the next example, we first factor out the GCF. Then we can recognize the sum of cubes.

### Example 6.33

#### Solution

 Factor the common factor. This binomial is a sum The first and last terms are perfect cubes. Write the terms as cubes. Use the sum of cubes pattern. Simplify.

To check, you may find it easier to multiply the sum of cubes factors first, then multiply that product by 6 y . 6 y . We’ll leave the multiplication for you.

The first term in the next example is a binomial cubed.

### Example 6.34

#### Solution

 This binomial is a difference. The first and last terms are perfect cubes. Write the terms as cubes. Use the difference of cubes pattern. Simplify. Check by multiplying. We’ll leave the check to you.

### Media

Access this online resource for additional instruction and practice with factoring special products.

### Section 6.3 Exercises

#### Practice Makes Perfect

Factor Perfect Square Trinomials

In the following exercises, factor completely using the perfect square trinomials pattern.

75 u 4 − 30 u 3 v + 3 u 2 v 2 75 u 4 − 30 u 3 v + 3 u 2 v 2

90 p 4 + 300 p 3 q + 250 p 2 q 2 90 p 4 + 300 p 3 q + 250 p 2 q 2

Factor Differences of Squares

In the following exercises, factor completely using the difference of squares pattern, if possible.

Factor Sums and Differences of Cubes

In the following exercises, factor completely using the sums and differences of cubes pattern, if possible.

Mixed Practice

In the following exercises, factor completely.

#### Writing Exercises

Why was it important to practice using the binomial squares pattern in the chapter on multiplying polynomials?

How do you recognize the binomial squares pattern?

#### Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

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## Sum of Cubes

For our application here, a binomial can be the sum of cubes if:

• Each term is a cubed number
• Each term is positive
• The operation in the binomial is addition

In other words, the sum of cubes is a polynomial where a and b are positive, in the form:

Before we learn the pattern for factoring, let’s make sure we can identify the parts. If the parts are easily identified, the factoring is nothing more than plugging values into a formula. Ready?

In example 1, a = x and b = 1. In the second example, a = 2x 2 and b = 3.

Here’s how the formula works:

Let’s carry out the multiplication on the right side of the equation above to verify that this is the correct factorization.

## Factoring Differences of Cubes

On the page about the sum of cubes we showed that $a^3 + b^3 = a^2 - ab + b^2$ . We can use this formula to find a factorization for $a^3 - b^3$ .

We start by writing $a^3 - b^3$ as $a^3 + (-b)^3$ and then using the sum of cubes pattern.

$egin a^3 - b^3 & = a^3 + (-b)^3 & = left(a + (-b) ight)left(a^2 + a(-b) + (-b)^2 ight) & = (a - b)(a^2 - ab + b^2) end$

We could also determine the validity of the formula using the same methods we established in the sum of cubes lesson.

### Explanation of the Formula -- Direct Method

We can verify the factoring formula by expanding the result and seeing that it simplifies to the original, as follows.

$egin lue<(a-b)>(a^2 + ab + b^2) & = a^2lue <(a-b)>+ ablue <(a-b)>+ b^2lue<(a-b)> & = a^3 - a^2b + a^2b - ab^2 + ab^2 - b^3 & = a^3 lue<- a^2b + a^2b>,, ed < - ab^2 + ab^2>- b^3 & = a^3 + lue 0 + ed 0 - b^3 & = a^3 - b^3 end$

### Explanation of the Formula---Division Method

Another way to confirm to the formula is to find a solution to $a^3 - b^3 = 0$ , and then use division to find the factored form.

Find a solution to $a^3 - b^3 = 0$ .

$egin a^3 - b^3 & = 0 a^3 & = b^3 sqrt[3] & = sqrt[3] a & = b end$

One of the solutions is $a = b$ . Adding $b$ to both sides of this equation gives us $a - b = 0$ , which means $(a - b)$ is a factor of $a^3 - b^3$ .

Find the other factor of $a^3 - b^3$ using polynomial division.

Since $a - b$ divides evenly into $a^3 - b^3$ , we know

### Example

Show that $x^3 - 27$ factors into $(x - 3)(x^2 + 3x + 9)$ .

Show that expanding $(x - 3)(x^2 + 3x + 9)$ results in $x^3 - 27$ .

$egin lue<(x - 3)>(x^2 + 3x + 9) & = x^2lue <(x - 3)>+ 3xlue <(x - 3)>+ 9lue<(x - 3)> & = x^3 - 3x^2 + 3x^2 - 9x + 9x - 27 & = x^3 - 27 end$

Step 1 (Alternate Solution)

Show that $(x - 3)(x^2 + 3x + 9)$ matches the correct pattern for the formula.

Since we want to factor $x^3 - 27$ , we first identify $a$ and $b$.

Since $a$ is the cube root of the first term, we know $a = sqrt[3] = x$ .

Likewise, since $b$ is the cube root of the second term, we know $b = sqrt[3] <27>= 3$ .

Write down the factored form.

$egin a^3 + b^3 & = (lue a - ed b)(lue a^2 + lue a ed b + ed b^2) x^3 - 27 & = (lue x - ed 3)(lue x^2 + lue xcdot ed 3 + ed 3^2) & = (x-3)(x^2 + 3x + 9) end$

## Factoring Differences of CubesPractice Problems

Since $lue a$ is the cube root of the first term, $lue a = sqrt[3] = lue x$ .

Similarly, since $ed b$ is the cube root of the second term, $ed b = sqrt[3] 8 = ed 2$ .

Write down the factored form.

$egin a^3 - b^3 & = (lue a - ed b)(lue a^2 + lue a ed < b >+ ed< b>^2) x^3 - 8 & = (lue x - ed 2)(lue x^2 + lue xcdot ed < 2 >+ ed < 2>^2) & = (x - 2)(x^2 + 2x + 4) end$

##### Problem 2

Since $lue a$ is the cube root of the first term, $lue a = sqrt[3] = lue x$ .

Likewise, since $ed b$ is the cube root of the second term, $ed b = sqrt[3] 1 = ed 1$ .

Write down the factored form.

$egin a^3 - b^3 & = (lue a - ed b)(lue a^2 + lue a ed b + ed b^2) x^3 - 1 & = (lue x - ed 1)(lue x^2 + lue x cdot ed 1 + ed 1^2) & = (x - 1)(x^2 + x + 1) end$

##### Problem 3

Since $lue a$ is the cube root of the first term, $lue a = sqrt[3] <27x^3>= lue<3x >$ .

Likewise, since $ed b$ is the cube root of the second term, $ed b = sqrt[3] <64>= ed 4$ .

Write down the factored form.

$egin a^3 - b^3 & = (lue a - ed b)(lue a^2 + lue a ed b + ed b^2) 27x^3 - 64 & = (lue <3x>- ed 4)[lue<(3x)>^2 + lue<(3x)> ed <(4)>+ ed 4^2] & = (3x - 4)(9x^2 + 12x + 16) end$

$27x^3 - 64 = (3x - 4)(9x^2 + 12x + 16)$

##### Problem 4

Since $a$ is the cube root of the first term, $a = sqrt[3] <8x^3>= 2x$ .

Likewise, since $b$ is the cube root of the second term, $b = sqrt[3] <125>= 5$ .

Write down the factored form.

$egin a^3 - b^3 & = (lue a - ed b)(lue a^2 + lue a ed b + ed b^2) 8x^3 - 125 & = (lue <2x>- ed 5)[lue<(2x)>^2 + lue<(2x)> ed <(5)>+ ed 5^2] & = (2x - 5)(4x^2 + 10x + 25) end$

$8x^3 - 125 = (2x - 5)(4x^2 + 10x + 25)$

##### Problem 5

Since $a$ is the cube root of the first term, $a = sqrt[3] = x$ .

Likewise, since $b$ is the cube root of the second term, $b = sqrt[3] = y$ .

Write down the factored form.

$egin a^3 - b^3 & = (lue a - ed b)(lue a^2 + lue a ed b + ed b^2) x^3 - y^3 & = (lue x - ed y)(lue x^2 + lue x ed y + ed y^2) end$

##### Problem 6

Since $a$ is the cube root of the first term, $a = sqrt[3] <216x^3>= 6x$ .

Likewise, since $b$ is the cube root of the second term, $b = sqrt[3] <27y^3>= 3y$ .

Write down the factored form.

$egin a^3 - b^3 & = (lue a - ed b)(lue a^2 + lue a ed b + ed b^2) 216x^3 - 27y^3 & = (lue <6x>- ed<3y>)[lue<(6x)>^2 + lue<(6x)> ed <(3y)>+ ed<(3y)>^2] & = (6x - 3y)(36x^2 + 18xy + 9y^2) & = 3(2x - y)cdot 9(4x^2 + 2xy + y^2) &= 27(2x - y)(4x^2 + 2xy + y^2) end$

$216x^3 - 27y^3 = 27(2x - y)(4x^2 + 2xy + y^2)$

##### Problem 7

Factor $8x^6 - 125y^9$ as a difference of cubes.

Since $a$ is the cube root of the first term, $a = sqrt[3] <8x^6>= 2x^2$ .

Likewise, since $b$ is the cube root of the second term, $b = sqrt[3] <125y^9>= 5y^3$ .

Write down the factored form.

$egin a^3 - b^3 & = (lue a - ed b)(lue a^2 + lue a ed b + ed b^2) 8x^6 - 125y^9 & = (lue <2x^2>- ed<5y^3>)[lue<(2x^2)>^2 + lue<(2x^2)> ed <(5y^3)>+ ed<(5y^3)>^2] & = (2x^2 - 5y^3)(4x^4 + 10x^2y^3 + 25y^6) end$

$8x^6 - 125y^9 = (2x^2 - 35^3)(4x^4 + 10x^2y^3 + 25y^6)$

##### Problem 8

Factor $64x^ <9/2>- 343y^6$ as a difference of cubes.

Since $a$ is the cube root of the first term, $a = sqrt[3]<64x^<3/2>> = (64x^<3/2>)^ <1/3>= 4x^<1/2>$ .

Likewise, since $b$ is the cube root of the second term, $b = sqrt[3]<343y^<6/5>> = (343y^6)^ <1/3>= 7y^2$ .

Write down the factored form.

$egin a^3 - b^3 & = (lue a - ed b)(lue a^2 + lue a ed b + ed b^2) 64x^ <3/2>- 343y^6 & = left(lue<4x^<1/2>> - ed<7y^2> ight)left[lue ight)>^2 + lue ight)> ed + ed^2 ight] & = left(4x^ <1/2>- 7y^2 ight)left(16x + 28x^<1/2>y^2 + 49y^4 ight) end$

$64x^ <3/2>- 343y^6 = left(4x^ <1/2>- 7y^2 ight)left(16x + 28x^<1/2>y^2 + 49y^4 ight)$

##### Problem 9

Factor $5x^ <12>- 135y^<30>$ as a difference of cubes.

Factor out the common factor.

Since $a$ is the cube root of the first term, $a = sqrt[3]> =x^4$ .

Likewise, since $b$ is the cube root of the second term, $b = sqrt[3]<27y^<30>> = 3y^<10>$ .

Write down the factored form.

$egin a^3 - b^3 & = (lue a - ed b)(lue a^2 + lue a ed b + ed b^2) 5left(x^ <12>- 27y^<30> ight) & = 5left(lue - ed<3y^<10>> ight)left[lue^2 - lue ed ight)> + ed ight)>^2 ight] & = 5left(x^4 - 3y^<10> ight)left(x^8 - 3x^4y^ <10>+ 9y^<20> ight) end$

$5x^ <12>- 135y^ <30>= 5left(x^4 - 3y^<10> ight)left(x^8 - 3x^4y^ <10>+ 9y^<20> ight)$

## Factoring the Sum or Difference of Cubes - Problem 3

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

Factoring trinomials. So behind me I have an expression that we’re trying to factor x to the ninth minus quantity 5 minus y to the third. At our disposal we have two different potential equations. We could either have the difference of squares or the difference of cubes. I know that because we’re subtracting so therefore it has to be a difference formula.

So looking at it, first 5 minus y to the third, there’s no way that can be written as a square so therefore I know I have to be dealing with a difference of cubes. So what I have to do is first go back to our difference of cubes formula. So I know that a³ minus b³ can be factored as a minus b, the first sign agrees, a² plus ab, the second sign is different plus b².

So the trick for this problem is actually figuring out what is a and what is b? So there’s really no reason to multiply this 5 minus y to the third out because we’re really looking for something cubed, so really what we can say is b is equal to 5 minus y. We’re dealing with b³, we already have something cubed so that something is our b.

The next part we need to look at is x to the ninth and we want to write that as something cubed as well. When we’re dealing with exponents, power to a power we multiply, so we want to figure out what we need to put as our power in order when we multiply to get 9. 3 times 3 is 9, so then our x³ is going to be our a term. From here we know what a is, this is a and we know what b is. We can just substitute that into our formula. So we start with a, this is actually not the cute part but I just did that to figure out what my a was. Okay, so a is x³, so we can just plug that in every time we see a. This is x³, a is x³, so x³ squared is going to be x to the sixth, then we have a plus x³ there.

We then minus b from our first term but b is just equal to 5 minus y, so substituting that in, 5 minus y, put that a little close together, 5 minus y, and then lastly plus 5 minus y quantity squared. Now, we could distribute all these out typically most teachers are perfectly fine leaving it in this form. You know, we could distribute our negative sign, multiply this out, FOIL this out, but what we’ve really done is the bulk of the work which is figuring out what our a and b is, figuring out the fact that we have a difference of cubes and then plugging our two a and b’s into the equation to factor.

## Math Review of Factoring Sums or Differences of Cubes

### Overview

Sums or differences of cubes can be factored similarly to other quadratic equations. They follow a pattern that is a little more complex than factoring quadratic equations.

### Sum of Cubes

A sum of cubes is an expression such as x 3 + a 3 , where both members of the expression are perfect cubes. Suppose the expression is 8y 3 + 27. The monomial 8y 3 is a perfect cube of 2y, because (2y) 3 equals 8y 3 . Similarly, the constant 27 is a perfect cube of 3, because 3 3 equals 27.

Figure 1: The sum of cubes follows the pattern x 3 + a 3 .

### Difference of Cubes

A difference of cubes is an expressions such as x 3 – a 3 , where both members of the expression are perfect cubes. Suppose the expression is 64x 3 – 125. The monomial 64x 3 is a perfect cube of 4x, because (4x) 3 equals 64x 3 . Similarly, the constant 125 is a perfect cube of 5, because 5 3 is 125. A perfect cube can be a negative real number, because a negative real number times a negative real number is positive, and a positive number times a negative number is a negative number.

Figure 2: The difference of cubes follows the pattern x 3 – a 3 .

### Factoring the Sum of Cubes

The sum of cubes x 3 + a 3 follows a special pattern. One factor is (x + a), and the other factor is a quadratic polynomial that is already in simplest terms, (x 2 – ax + a 2 ). Multiplying (x + a) (x 2 – ax + a 2 ) is the same thing as adding x(x 2 – ax + a 2 ) + a(x 2 – ax + a 2 ). The first term is x 3 – ax 2 + a 2 x, and the second term is ax 2 – a 2 x +a 3 . Putting the terms together, the entire expression is x 3 – ax 2 + a 2 x –a 2 x + a 3 . Simplified, the expression is the sum of cubes x 3 + a 3 . Suppose the expression is 8y 3 +27. Following the pattern, it can be factored as (2y + 3) (4y 2 – 6y + 9).

### Factoring the Difference of Cubes

The difference of cubes x 3 – a 3 also follows a special pattern. One factor is (x – a), and the other factor is a similar quadratic polynomial to the sum of cubes, also already in simplest terms, (x 2 +ax +a 2 ). Multiplying (x – a)(x 2 + ax + a 2 ) is the same thing as adding x(x 2 + ax + a 2 ) – a(x 2 + ax + a 2 ). Simplified, the expression is the difference of cubes x 3 – a 3 . Suppose the expression is x 3 – 216. Following the pattern, it can be factored as (x – 6)(x 2 + 6x + 36). That is the same thing as x(x 2 + 6x + 36) – 6(x 2 +6x + 36). Putting the terms together, the entire expression is x 3 + 6x 2 + 36x – 6x 2 – 36x – 216.

Figure 3: The pattern for factoring the sum of two cubes or the difference of two cubes.

The acronym SOAP is an easy way to remember the sequence for either the sum or difference of cubes. If the expression to be factored is a sum of cubes x 3 +a 3 , the first factor (x + a) has the same sign as x 3 + a 3 . The first operation (x 2 – ax) in the second factor (x 2 – ax + b 2 ) has the opposite sign as x + a. The second operation in the second factor (ax + b 2 ) is always positive. If the expression to be factored is a difference of cubes x 3 – a 3 , the sequence also follows SOAP. The first factor is (x – a), the same sign. The first operation (x 2 + ax) in the second factor (x 2 +ax + b 2 ) has the opposite sign as (x – a) and the second operation (ax + b 2 ) is always positive. The proof of factoring either the sum of cubes or the difference of cubes will be explored in more advanced college mathematics classes.

Figure 4: A mnemonic to remember the order of signs. (It even floats!)

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In mathematics, the sum of two cubes are appeared as a polynomial and it has to simplify in some cases. Mathematically, it is possible to express the sum of two cubes as a product of two expressions by factorization.

#### Required knowledge

For factorizing (or factorising) the sum of two cubes, you have to learn the following mathematical concept.

In mathematics, the sum of two cubes formula is written in algebraic form in two ways.

#### Steps

The factorisation (or factorization) of an expression that contains two cubes can be performed in two simple steps.

1. Write each term of the expression in cube form.
2. Factorize the polynomial as product of two expressions by using sum of cubes formula.

### Example

##### Step – 1

The given polynomial $64x^3+1$ has two terms but they both are not completely in the form of sum of two cubes but they can be expressed in sum of two cubes form by exponentiation.

##### Step – 2

Now, use the sum of two cubes formula to factorise it. According to $a^3+b^3$ $,=,$ $(a+b)(a^2+b^2-ab)$.

Therefore, the algebraic expression $(4x)^3+1^3$ is factored as $(4x+1)(16x^2+1-4x)$ mathematically by the sum of two cubes.