What Is A Difference Of Cubes? (3 Key Ideas You Should Know)

A difference of cubes is used in algebra to factor cubic or higher degree polynomials.  The predictable formula for factoring a difference of cubes saves us time if we can recognize the general form.

So, what is a difference of cubes?  A difference of cubes has the form A³ – B³ and factors as (A – B)(A² + AB + B²). Each term in a difference of cubes has a perfect cube numerical coefficient and exponents that are multiples of 3. In some cases, we need to factor out a GCF (greatest common factor) to reveal a difference of cubes.

Of course, there is also a formula for a sum of cubes: A³ + B³ = (A + B)(A² – AB + B²).

In this article, we’ll talk about difference of cubes and when to use this formula.  We’ll also answer some common questions and look at some examples to make the concepts clear.

Let’s get started.

What Is A Difference Of Cubes?

A difference of cubes is an expression with exactly two terms with opposite signs (one is positive, and the other is negative). Each term has a perfect cube numerical coefficient (such as 1, 8, 27, 64, 125, etc.), and the exponent of every variable is a multiple of 3 (3, 6, 9, 12, etc.)

The formula for factoring a difference of cubes is given by:

• A³ – B³ = (A – B)(A² + AB + B²)

The following table summarizes what you should look for to recognize a difference of cubes.

Characteristic	What To Look For
Number Of Terms	2
Signs Of Terms	+ and – (one positive, one negative)
Numerical Coefficients	Perfect Cubes: 1, 8, 27, 64, 125, 216, etc.
Exponents On Variables	Multiples of 3: 3, 6, 9, 12, 15, 18, 21 etc.

Note that we can also write a similar formula for factoring a sum of cubes:

• A³ + B³ = (A + B)(A² – AB + B²)

When To Use Difference Of Cubes

We use a difference of cubes to factor certain cubic expressions or as part of the process to factor higher-degree polynomials.

Let’s take a look at some examples that show how to factor a difference of cubes.

Example 1: Factor A Difference Of Cubes

Let’s say we want to factor the expression 8x³ – y⁶.

We can recognize this as a difference of cubes, since:

• There are exactly two terms: 8x³ and –y³
• The two terms have opposite signs (+ and –)
• The numerical coefficients are perfect cubes: 8 = 2³ and 1 = 1³
• The exponents on the variables are multiples of 3: 3 for x³ and 6 for y⁶

So, we can factor this expression using a difference of cubes.  To find A and B, we solve:

• A³ = 8x³
• (A³)^1/3 = (8x³)^1/3
• A = 8^1/3x
• A = 2x

Similarly:

• B³ = y⁶
• (B³)^1/3 = (y⁶)^1/3
• B = y²

Now, we substitute A = 2x and B = y into the formula for factoring a difference of cubes:

• A³ – B³ = (A – B)(A² + AB + B²)
• (2x)³ – (y²)³ = (2x – y²)((2x)² + (2x)(y²) + (y²)²)
• 8x³ – y⁶ = (2x – y²)(4x² + 2xy² + y⁴)

Example 2: Factor A Difference Of Cubes

Let’s say we want to factor the expression 27x⁶y⁹ – 64s³t¹²

We can recognize this as a difference of cubes, since:

• There are exactly two terms: 27x⁶y⁹ and – 64s³t¹²
• The two terms have opposite signs (+ and –)
• The numerical coefficients are perfect cubes: 27 = 3³ and 64 = 4³
• The exponents on the variables are multiples of 3: 6 for x⁶, 9 for y⁹, 3 for s³, and 12 for t¹².

In this case, we have A = 3x²y³ and B = 4st⁴.

Now, we substitute A = 3x²y³ and B = 4st⁴ into the formula for factoring a difference of cubes:

• A³ – B³ = (A – B)(A² + AB + B²)
• (3x²y³)³ – (4st⁴)³ = (3x²y³ – 4st⁴)((3x²y³)² + (3x²y³)(4st⁴) + (4st⁴)²)
• 27x⁶y⁹ – 64s³t¹² = (3x²y³ – 4st⁴)(9x⁴y⁶ + 12x²y³st⁴ + 16s²t⁸)

Example 3: Factor A Difference Of Cubes

Let’s say we want to factor the expression 250x⁴y⁸z – 16xy²z¹⁰.

This does not look like a difference of cubes.  However, we can factor out a GCF (greatest common factor) of 2xy²z from both terms to get:

• 250x⁴y⁸z – 16xy²z¹⁰ = 2xy²z(125x³y⁶ – 8z⁹)

Now we can recognize the expression in parentheses as a difference of cubes, since:

• There are exactly two terms: 125x³y⁶ and – 8z⁹
• The two terms have opposite signs (+ and –)
• The numerical coefficients are perfect cubes: 125 = 5³ and 8 = 2³
• The exponents on the variables are multiples of 3: 3 for x³, 6 for y⁶, and 9 for z⁹.

In this case, we have A = 5xy² and B = 2z³.

Now, we substitute A = 5xy² and B = 2z³ into the formula for factoring a difference of cubes:

• A³ – B³ = (A – B)(A² + AB + B²)
• (5xy²)³ – (2z³)³ = (5xy²  – 2z³)((5xy²)² + (5xy²)( 2z³) + (2z³)²)
• 125x³y⁶ – 8z⁹ = (5xy²  – 2z³)(25x²y⁴ + 10xy²z³ + 4z⁶)

Don’t forget to include the GCF that we factored out earlier to factor completely:

• 250x⁴y⁸z – 16xy²z¹⁰
• = 2xy²z(125x³y⁶ – 8z⁹)  [factor out GCF]
• = 2xy²z(5xy²  – 2z³)(25x²y⁴ + 10xy²z³ + 4z⁶)  [factor difference of cubes]

Example 4: Factor A Sum Of Cubes

Let’s say we want to factor the expression x²⁷ – y²⁷.

We can recognize this as a difference of cubes, since:

• There are exactly two terms: x²⁷ and –y²⁷.
• The two terms have opposite signs (+ and –)
• The numerical coefficients are perfect cubes: 1 = 1³ and 1 = 1³
• The exponents on the variables are multiples of 3: 27 for x²⁷ and 27 for y²⁷.

In this case, we have A = x⁹ and B = y⁹.

Now, we substitute A = x⁹ and B = y⁹ into the formula for factoring a difference of cubes:

• A³ – B³ = (A – B)(A² + AB + B²)
• (x⁹)³ – (y⁹)³ = (x⁹ – y⁹)((x⁹)² + (x⁹)( y⁹) + (y⁹)²)
• x²⁷ – y²⁷ = (x⁹ – y⁹)(x¹⁸ + x⁹y⁹ + y¹⁸)

However, this is not factored completely.  Note that the first factor, x⁹ – y⁹, is also a difference of cubes, so we factor it as well:

• x²⁷ – y²⁷
• =(x⁹ – y⁹)(x¹⁸ + x⁹y⁹ + y¹⁸)
• =(x³ – y³)(x⁶ + x³y³ + y⁶)(x¹⁸ + x⁹y⁹ + y¹⁸)

However, we are still not done, since x³ – y³ is also a difference of cubes:

• =(x – y)(x² + xy + y²)(x⁶ + x³y³ + y⁶)(x¹⁸ + x⁹y⁹ + y¹⁸)

So, the complete factorization (after using a difference of cubes three times) is:

• x²⁷ – y²⁷ = (x – y)(x² + xy + y²)(x⁶ + x³y³ + y⁶)(x¹⁸ + x⁹y⁹ + y¹⁸)

Example 5: Factor A Sum Of Cubes

Let’s say we want to factor the expression 125x⁶ + y⁹.

We can recognize this as a sum of cubes, since:

• There are exactly two terms: 125x⁶ and y⁹
• The two terms have the same signs (+ and +)
• The numerical coefficients are perfect cubes: 125 = 5³ and 1 = 1³
• The exponents on the variables are multiples of 3: 6 for x⁶ and 9 for y⁹

So, we can factor this expression using a sum of cubes.  In this case, A = 5x² and B = y³.

Now, we substitute A = 5x² and B = y³ into the formula for factoring a sum of cubes:

• A³ + B³ = (A + B)(A² – AB + B²)
• (5x²)³ + (y³)³ = (5x² + y³)((5x²)² – (5x²)(y³) + (y³)²)
• (125x⁶ + y⁹) = (5x² + y³)(25x⁴ – 5x²y³ + y⁶)

Example 6: Factor A Sum Of Cubes

Let’s say we want to factor the expression 40x⁸y³z + 135x²z¹⁰.

This does not look like a sum of cubes.  However, we can factor out a GCF (greatest common factor) of 5x²z from both terms to get:

• 40x⁸y³z + 135x²z¹⁰ = 5x²z(8x⁶y³ + 27z⁹)

Now we can recognize the expression in parentheses as a sum of cubes, since:

• There are exactly two terms: 8x⁶y³ and 27z⁹
• The two terms have the same signs (+ and +)
• The numerical coefficients are perfect cubes: 8 = 2³ and 27 = 3³
• The exponents on the variables are multiples of 3: 6 for x⁶, 3 for y³, and 9 for z⁹.

In this case, we have A = 2x²y and B = 3z³.

Now, we substitute A = 2x²y and B = 3z³ into the formula for factoring a sum of cubes:

• A³ + B³ = (A + B)(A² – AB + B²)
• (2x²y)³ + (3z³)³ = (2x²y + 3z³)((2x²y)² – (2x²y)( 3z³) + (3z³)²)
• 8x⁶y³ + 27z⁹ = (2x²y + 3z³)(4x⁴y² – 6x²yz³ + 9z⁶)

How To Derive The Difference Of Cubes Formula

We can derive the sum of cubes formula if we use the Distributive Property and start with the right side of the equation:

• (A – B)(A² + AB + B²)
• =A(A² + AB + B²) – B(A² + AB + B²)  [distribute (A² + AB + B²) through first set of parentheses]
• =(A³ + A²B + AB²) – (BA² + AB² + B³) [distribute A through 1st parentheses, then distribute B through 2^nd parentheses]
• =(A³ + A²B + AB²) – BA² – AB² – B³  [distribute negative sign through 2^nd parentheses]
• =A³ + A²B – BA² – B³  [AB² – AB² = 0]
• =A³ – B³  [A²B – BA² = 0]

So, this proves that the formula for a difference of cubes is correct.

We can use similar algebra to prove the formula for a sum of cubes as well.  A shorter approach is to replace “B” with “-B” in the difference of cubes formula to get:

• (A³ – B³) = (A – B)(A² + AB + B²)  [difference of cubes formula]
• (A³ – (-B)³) = (A – (-B))(A² + A(-B) + (-B)²)  [replace B by –B everywhere in the formula]
• (A³ – (-B³) = (A + B)(A² – AB + B²)  [(-B)² = B² and (-B)³ = -B³]
• (A³ + B³) = (A + B)(A² – AB + B²) 

So, this proves that the formula for a sum of cubes is correct.

Conclusion

Now you know what a difference of cubes is, when to use it, and how to apply the formula.  You also know how to apply a sum of cubes in the same way.

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