Examples of Perfect Cubes

Here are a couple of examples of perfect cubes:

3³ = 27 and 2³ = 8

As you can see from the examples, perfect cubes are the third power of a number or variable. So, you get a perfect cube when you multiply a number by itself and then multiply that answer again by the first number.

Perfect Cubes – Exercises

Instructions: Factor the following algebraic expressions. To get the shortcut form for your answers, please refer to the examples below this exercise.

[WpProQuiz 13]

Perfect Cubes and Binomials

The difference of two perfect cubes is a binomial.

You will remember that a binomial is an algebraic expression that has two terms.

Here is one of our examples of an expression that shows the difference of two perfect cubes:

x³ – y³

The sum of two perfect cubes is also a binomial.

For example, we can change the expression above to show the sum of perfect cubes by using the plus sign.

x³ + y³

Difference of Two Perfect Cubes

You will need to know how to factor the difference of perfect cubes on your examination.

The second of our examples of an algebraic expression for the difference of perfect cubes is shown below.

x³ – y³

The form for factoring the difference of perfect cubes is as follows:

x³ – y³ =

(x – y)(x² + xy + y²)

You should memorize the above form for your exam.

Proof of Difference of Two Cubes

We can prove the above form by applying the distributive property of multiplication.

Multiply the first term in the first set of parentheses by all of the terms in the second set of parentheses.

x³ – y³ =

(x – y)(x² + xy + y²) =

[(x × x²) + (x × xy) + (x × y²)] – [(y) × (x² + xy + y²)] =

[(x × x²) + (x × xy) + (x × y²)] + [(–y) × (x² + xy + y²)] =

[(x³) + (x²y) + (xy²)] + [(–y) × (x² + xy + y²)] =

(x³ + x²y + xy²) × [(–y) + (x² + xy + y²)]

Then multiply the second term in the first set of parentheses by all of the terms in the second set of parentheses.

Remember to be careful with the plus or minus sign.

We are subtracting the terms in the first set of parentheses, so we need to multiply all of the terms in the second set of parentheses by (–y).

(x³ + x²y + xy²) + [(–y) × (x² + xy + y²)] =

(x³ + x²y + xy²) + [(–y × x²) + (–y × xy) + (–y × y²)] =

(x³ + x²y + xy²) + [(–yx²) + (–xy²) + (–y³)] =

(x³ + x²y + xy²) + [(–x²y) + (–xy²) + (–y³]) =

(x³ + x²y + xy²) + (–x²y + –xy² + –y³)

Then group like terms together and simplify to finish the proof.

(x³ + x²y + xy²) + (–x²y + –xy² + –y³) =

x³ + x²y + xy² – x²y – xy² – y³ =

x³ + x²y – x²y + xy² – xy² – y³ =

x³ ~~+ x²y – x²y~~ ~~+ xy² – xy²~~ – y³ =

x³ – y³

Sum of Two Perfect Cubes

You will need to know how to factor the sum of perfect cubes for your math test.

An algebraic expression for the sum of perfect cubes is as follows:

x³ + y³

The form for factoring the sum of perfect cubes is:

x³ + y³ =

(x + y)(x² – xy + y²)

You should also know the above above form by heart for your math test.

Proof of Sum of Two Cubes

We can also prove the above form for the sum of perfect cubes with the distributive property of multiplication.

Remember to multiply the first term in the first set of parentheses by all of the terms in the second set of parentheses.

x³ + y³ =

(x + y)(x² – xy + y²) =

[(x × x²) + (x × –xy) + (x × y²)] + [(y) × (x² – xy + y²)] =

[(x³) + (–x²y) + (xy²)] + [(y) × (x² – xy + y²)] =

(x³ + –x²y + xy²) × [(y) + (x² – xy + y²)]

Then multiply the second term in the first set of parentheses by all of the terms in the second set of parentheses.

(x³ + –x²y + xy²) + [(y) × (x² – xy + y²)] =

(x³ – x²y + xy²) + [(y × x²) + (y × –xy) + (y × y²)] =

(x³ – x²y + xy²) + [(yx²) + (–xy²) + (y³)] =

(x³ – x²y + xy²) + (x²y + –xy² + y³)

Then group like terms together and simplify to finish the proof.

(x³ – x²y + xy²) + (x²y + –xy² + y³) =

x³ – x²y + xy² + x²y – xy² + y³ =

x³ – x²y + x²y + xy² – xy² + y³ =

x³ ~~– x²y + x²y~~ ~~+ xy² – xy²~~ + y³ =

x³ + y³

You may also want to look at our practice problems on factoring perfect squares.

You should also try our free practice problems on factoring and algebra.

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