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## Sum or Difference of Cubes

## Sum and Difference of Cubes

The sum or difference of two cubes can be factored into a product of a binomial times a trinomial.

A mnemonic for the signs of the factorization is the word “SOAP”, the letters stand for “Same sign” as in the middle of the original expression, “Opposite sign”, and “Always Positive”.

## Sums and Differences of Cubes

The other two special factoring formulas you’ll need to memorize are very similar to one another; they’re the formulas for factoring the sums and the differences of cubes. Here are the two formulas:

Factoring a Sum of Cubes:

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

Factoring a Difference of Cubes:

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

You’ll learn in more advanced classes how they came up with these formulas. For now, just memorize them.

To help with the memorization, first notice that the *terms* in each of the two factorization formulas are exactly the same. Then notice that each formula has only one “minus” sign. The distinction between the two formulas is in the location of that one “minus” sign:

For the *difference* of cubes, the “minus” sign goes in the linear factor, *a* – *b*; for the *sum* of cubes, the “minus” sign goes in the quadratic factor, *a*^{2} – *ab* + *b*^{2}.

Some people use the mnemonic “**SOAP**” to help keep track of the signs; the letters stand for the linear factor having the “same” sign as the sign in the middle of the original expression, then the quadratic factor starting with the “opposite” sign from what was in the original expression, and finally the second sign inside the quadratic factor is “always positive”.

*a*^{3} ± *b*^{3} = (*a* [**Same** sign] *b*)(*a*^{2} [**Opposite** sign] *ab* [**Always** **Positive**] *b*^{2})

Whatever method best helps you keep these formulas straight, use it, because you should not assume that you’ll be given these formulas on the test. You should expect to need to know them.

Note: The quadratic portion of each cube formula *does not factor*, so don’t waste time attempting to factor it. Yes, *a*^{2} – 2*ab* + *b*^{2} and *a*^{2}+ 2*ab* + *b*^{2} factor, but that’s because of the 2’s on their middle terms. These sum- and difference-of-cubes formulas’ quadratic terms *do not have* that “2”, and thus *cannot* factor.

When you’re given a pair of cubes to factor, carefully apply the appropriate rule. By “carefully”, I mean “using parentheses to keep track of everything, especially the negative signs”. Here are some typical problems:

- Factor
*x*^{3}– 8

This is equivalent to *x*^{3} – 2^{3}. With the “minus” sign in the middle, this is a difference of cubes. To do the factoring, I’ll be plugging *x* and 2 into the difference-of-cubes formula. Doing so, I get:

*x*^{3} – 8 = *x*^{3} – 2^{3}

= (*x* – 2)(*x*^{2} + 2*x* + 2^{2})

= **( x – 2)(x^{2} + 2x + 4)**

- Factor 27
*x*^{3}+ 1

The first term contains the cube of 3 and the cube of *x*. But what about the second term?

Before panicking about the lack of an apparent cube, I remember that 1 can be regarded as having been raised to any power I like, since 1 to any power is still just 1. In this case the power I’d like is 3, since this will give me a sum of cubes. This means that the expression they’ve given me can be expressed as:

(3*x*)^{3} + 1^{3}

So, to factor, I’ll be plugging 3*x* and 1 into the sum-of-cubes formula. This gives me:

27*x*^{3} + 1 = (3*x*)^{3} + 1^{3}

= (3*x* + 1)((3*x*)^{2} – (3*x*)(1) + 1^{2})

= **(3 x + 1)(9x^{2} – 3x + 1)**

- Factor
*x*^{3}*y*^{6}– 64

First, I note that they’ve given me a binomial (a two-term polynomial) and that the power on the *x* in the first term is 3 so, even if I weren’t working in the “sums and differences of cubes” section of my textbook, I’d be on notice that maybe I should be thinking in terms of those formulas.

Looking at the other variable, I note that a power of 6 is the cube of a power of 2, so the other variable in the first term can be expressed in terms of cubing, too; namely, as the cube of the square of *y*.

The second term is 64, which I remember is the cube of 4. (If I didn’t remember, or if I hadn’t been certain, I’d have grabbed my calculator and tried cubing stuff until I got the right value, or else I’d have taken the cube root of 64.)

So I now know that, with the “minus” in the middle, this is a difference of two cubes; namely, this is:

(*xy*^{2})^{3} – 4^{3}

Plugging into the appropriate formula, I get:

*x*^{3}*y*^{6} – 64 = (*xy*^{2})^{3} – 4^{3}

= (*xy*^{2} – 4)((*xy*^{2})^{2} + (*xy*^{2})(4) + 4^{2})

= **( xy^{2} – 4)(x^{2}y^{4} + 4xy^{2} + 16)**

- Using an appropriate formula, factor 16
*x*^{3}– 250.

Um… I know that 16 is *not* a cube of anything; it’s actually equal to 2^{4}. What’s up?

What’s up is that they expect me to use what I’ve learned about simple factoring to first convert this to a difference of cubes. Yes, 16 = 2^{4}, but 8 = 2^{3}, a cube. I can get 8 from 16 by dividing by 2. What happens if I divide 250 by 2? I get 125, which is the cube of 5. So what they’ve given me can be restated as:

2(2^{3}*x*^{3} – 5^{3})

I can apply the difference-of-cubes formula to what’s inside the parentheses:

2^{3}*x*^{3} – 5^{3} = (2*x*)^{3} – (5)^{3}

= (2*x* – 5)((2*x*)^{2} + (2*x*)(5) + (5)^{2})

= (2*x* – 5)(4*x*^{2} + 10*x* + 25)

Putting it all together, I get a final factored form of:

2(2*x* – 5)(4*x*^{2} + 10*x* + 25)

- Factor –
*x*^{3}– 125.

My first reaction might be to go straight to applying the difference-of-cubes formula, since 125 = 5^{3}. But what about that “minus” sign in front?

Since neither of the factoring formulas they’ve given me includes a “minus” in front, maybe I can factor the “minus” out…?

–*x*^{3} – 125 = –1*x*^{3} – 125

= –1(*x*^{3} + 125)

Aha! Now what’s inside the parentheses is a *sum* of cubes, which I can factor. I’ve got the sum of the cube of *x* and the cube of 5, so:

*x*^{3} + 5^{3} = (*x* + 5)((*x*)^{2} – (*x*)(5) + (5)^{2})

= (*x* + 5)(*x*^{2} – 5*x* + 25)

Putting it all together, I get:

–1(*x* + 5)(*x*^{2} – 5*x* + 25)

## Difference of Cubes Formula

The difference of cubes formula is used to find the difference of cubes of two numbers without actually calculating the cubes. It is one of the algebraic identities. The difference of cubes formula is used to factorize the binomials of cubes. The difference of cubes formula is also known as a^{3 }– b^{3} formula. The difference of cubes formula is explained below along with solved examples in the following section.

## What Is the Difference of Cubes Formula?

The difference of cubes formula or a^{3 }– b^{3} formula can be verified, by multiplying (a – b) (a^{2} + ab + b^{2}) and see whether you get a^{3} – b^{3}.The difference of cubes formula is given as,

### Difference of Cubes Formula

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

You can remember these signs using the following trick.

## Examples Using Difference of Cubes Formula

**Example 1: **Find the value of 108^{3} – 8^{3 }by using the difference of cubes formula.

**Solution:**

To find: 108^{3} – 8^{3}.

Let us assume that a = 108 and b = 8.

We will substitute these in the formula of difference of cubes.

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

108^{3}−8^{3 }= (108 − 8)(108^{2 }+ (108)(8) + 8^{2}) = (100)(11664 + 864 + 64) = (100)(12592) = 1259200

**Answer:** 108^{3} – 8^{3} = 1,259,200.

**Example 2: **Factorize the expression 27x^{3} – 125 by using the difference of cubes formula.

**Solution:**

To factorize: 27x^{3} – 125.

We will use the difference of cubes^{ }formula to factorize this.

We can write the given expression as

27x^{3} – 125 = (3x)^{3} – 5^{3}

We will substitute a = 3x and b = 5 in the formula of the difference of cubes.

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

(3x)^{3 }− 5^{3 }= (3x − 5)((3x)^{2 }+ (3x)(5) + 5^{2}) = (3x − 5)(9x^{2} + 15x + 25)

**Answer: **27x^{3} – 125 = (3x – 5)(9x^{2} + 15x + 25).

**Difference of Two Cubes Formula**

An expression that occurs in the difference of two cubes usually is very hard to spot. The difference between the two cubes is equal to the difference of their cube roots, which contains the cube roots’ squares and the opposite of the cube roots’ product.

To see what distribution results in the difference of two cubes formula, we try to see if the distribution has a binomial,

(a – b),

which is the difference between two terms

(a² + ab + b²)

which has the opposite of their product and the squares of the two terms . Therefore the formula for the difference of two cubes is –

a³ – b³ = (a – b) (a² + ab + b²)

**Factoring Cubes Formula**

We always discuss the sum of two cubes and the difference of two cubes side-by-side. The idea is that they are related to formation. The only solution is to remember the patterns involved in the formulas.

Lets say –

Factoring x³ – 8,

This is equivalent to x³ – 2³**. **As the – sign is in the middle, it transpires into a difference of cubes**. **To do the factoring, so plugging x and 2 into the difference-of-cubes formula. Doing so, we get:

x³ – 8 = x³ – 2³

= (x – 2)(x² + 2x + 2²)

= (x – 2)(x² + 2x + 4)

## FAQs (Frequently Asked Questions)

**Q1. What is the Factoring Cubes Formula?**

Answer : The difference between the two cube formulas is in the area of the minus sign: For the difference of cubes, the – sign operates in the linear factor, a – b; for the sum of cubes, the – sign works in the quadratic factor, a²- ab + b².

**Q2. What Steps are Needed for Factoring a Sum of Cubes?**

Answer : To factor the sum of two cubes, determine if the two terms have the greatest common factor or GCF. Next, rewrite the first problem as a difference of two perfect cubes, then use those three pieces to write the final answer.

**Q3. Explain How an Expression is a Sum of Cubes?**

Answer : An expression requires two criteria to be factored as a sum of cubes. First, each term must be a cube. Secondly, each term should have the same sign, usually both positive.

## How to Break Down a Cubic Difference or Sum

After you’ve checked to see if there’s a Greatest Common Factor (GCF) in a given polynomial and discovered it’s a binomial that isn’t a difference of squares, you should consider that it may be a difference or sum of cubes.

A *difference of cubes* sounds an awful lot like a difference of squares, but it factors quite differently. A difference of cubes is a binomial that is of the form (something)^{3} – (something else)^{3}. To factor any difference of cubes, you use the formula *a*^{3} – *b*^{3} = (*a* – *b*)(*a*^{2} + *ab* + *b*^{2}).

A *sum of cubes* is a binomial of the form: (something)^{3} + (something else)^{3}. When you recognize a sum of cubes *a*^{3} + *b*^{3}, it factors as (*a* + *b*)(*a*^{2} – *ab* + *b*^{2}).

For example, to factor 8*x*^{3} + 27, you first look for the GCF. You find none, so now you use the following steps:

- Check to see if the expression is a difference of squares.You want to consider the possibility because the expression has two terms, but the plus sign between the two terms quickly tells you that it isn’t a difference of squares.
- Determine if you must use a sum or difference of cubes.The plus sign tells you that it may be a sum of cubes, but that clue isn’t foolproof. Time for some trial and error: Try to rewrite the expression as the sum of cubes; if you try (2
*x*)^{3}+ (3)^{3}, you’ve found a winner. - Break down the sum or difference of cubes by using the factoring shortcut.Replace
*a*with 2*x*and*b*with 3. The formula becomes [(2*x*) + (3)] [(2*x*)^{2}– (2*x*)(3) + (3)^{2}]. - Simplify the factoring formula.This example simplifies to (2
*x*+ 3)(4*x*^{2}– 6*x*+ 9). - Check the factored polynomial to see if it will factor again.You’re not done factoring until you’re done. Always look at the “leftovers” to see if they’ll factor again. Sometimes the binomial term may factor again as the difference of squares. However, the trinomial factor
*never*factors again.In this example, the binomial term 2*x*+ 3 is a first-degree binomial (the exponent on the variable is 1) without a GCF, so it won’t factor again. Therefore, (2*x*+ 3)(4*x*^{2}– 6*x*+ 9) is your final answer.

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