Difference of cubes formulas, equations, examples, exercises

The difference of cubes is a binomial algebraic expression of the form a³ - b³, where the terms a and b can be real numbers or algebraic expressions of various types. An example of a difference of cubes is: 8 - x³, since 8 can be written as 2³.

Geometrically we can think of a large cube with side a, from which the small cube with side b is subtracted, as illustrated in figure 1:

The volume of the resulting figure is precisely a difference of cubes:

V = a³ - b³

To find an alternative expression, it is observed that this figure can be decomposed into three prisms, as shown below:

A prism has a volume given by the product of its three dimensions: width x height x depth. In this way, the resulting volume is:

V = a³ - b³ = a^two.b + b³ + a.b^two

The factor b it is common to the right. Furthermore, in the figure shown above, it is particularly true that:

b = (a / 2) ⇒ a = b + b

Therefore it can be said that: b = a - b. Thus:

to³ - b³ = b (a^two + b^two +a.b) = (a-b) (a^two + a.b + b^two)

This way of expressing the difference of cubes will prove to be very useful in many applications and would have been obtained in the same way, even if the side of the missing cube in the corner was different from b = a / 2.

Note that the second parenthesislooks a lot like the remarkable product of the square of the sum, but the cross term is not multiplied by 2. The reader can develop the right side to verify that it is actually obtained to³ - b³.

Article index

• 1 Examples
  • 1.1 Factoring a difference of cubes
• 2 Exercise resolved
  • 2.1 Exercise 1
  • 2.2 Exercise 2
• 3 References

Examples

There are several differences of cubes:

1 - m⁶

to⁶b³ - 8z¹²Y⁶

(1/125) .x⁶ - 27 and⁹

Let's analice each one of them. In the first example, the 1 can be written as 1 = 1³ and the term m⁶ remains: (m^two)³. Both terms are perfect cubes, therefore their difference is:

1 - m⁶ = 1³ - (m^two)³

In the second example the terms are rewritten:

to⁶b³ = (a^twob)³

8z¹²Y⁶ = 2³ (z⁴)³ (Y^two)³ = (2z⁴Y^two)³

The difference of these cubes is: (a^twob)³ - (2z⁴Y^two)³.

Finally, the fraction (1/125) is (1/5³), x⁶ = (x^two)³, 27 = 3³ and and⁹ = (and³)³. Substituting all this in the original expression, you get:

(1/125) .x⁶  - 27y⁹ = [(1/5) (x^two)]³ - (3y³)³

Factoring a difference of cubes

Factoring the difference of cubes simplifies many algebraic operations. To do this, it is enough to use the formula deduced above:

Now, the procedure to apply this formula consists of three steps:

- First, the cube root of each of the terms of the difference is obtained.

- Then the binomial and the trinomial that appear on the right side of the formula are constructed.

- Finally, the binomial and the trinomial are substituted to obtain the definitive factorization.

Let's illustrate the use of these steps with each of the cube difference examples proposed above and thus obtain its factored equivalent.

Example 1

Factor the expression 1 - m⁶   following the steps outlined. We start by rewriting the expression as 1 - m⁶ = 1³ - (m^two)³ to extract the respective cube roots of each term:

Next, the binomial and the trinomial are constructed:

a = 1

b = m^two

Then:

a - b = 1 - m^two

 (to^two +a.b + b^two) = 1^two + 1.m^two + (m^two)^two = 1 + m^two + m⁴

 Finally it is substituted in the formula a³ - b³ = (a-b) (a^two +a.b + b^two):

1 - m⁶ = (1 - m^two) (1 + m^two + m⁴)

Example 2

Factorize:

to⁶b³ -8z¹²Y⁶ = (a^twob)³ - (2z⁴Y^two)³

Since these are perfect cubes, the cube roots are immediate: a^twob and 2z⁴Y^two, hence it follows that:

- Binomial: a^twob - 2z⁴Y^two

- Trinomial: (a^twob)^two + to^twob. 2z⁴Y^two + (to^twob + 2z⁴Y^two)^two

 And now the desired factorization is constructed:

to⁶b³ -8z¹²Y⁶ = (a^twob - 2z⁴Y^two). [(to^twob)^two + to^twob. 2z⁴Y^two + (to^twob + 2z⁴Y^two)^two] =

= (a^twob - 2z⁴Y^two). [to⁴b^two + 2nd^twob.z⁴Y^two + (to^twob + 2z⁴Y^two)^two]

In principle, the factoring is ready, but it is often necessary to simplify each term. Then we develop the remarkable product -square of a sum- that appears at the end and then add like terms. Remembering that the square of a sum is:

(x + y)^two = x^two + 2xy + and^two

The notable product on the right is developed like this:

(to^twob + 2z⁴Y^two)^two = a⁴b^two + 4th^twob.z⁴Y^two + 4z⁸Y⁴

 Substituting the expansion obtained in the factorization of the difference of cubes:

to⁶b³ -8z¹²Y⁶ = (a^twob - 2z⁴Y^two). [to⁴b^two + 2nd^twob.z⁴Y^two + to⁴b^two + 4th^twob.z⁴Y^two + 4z⁸Y⁴] =

Finally, grouping like terms and factoring the numerical coefficients, which are all even, we obtain:

(to^twob - 2z⁴Y^two). [2a⁴b^two + 6th^twob.z⁴Y^two + 4z⁸Y⁴] = 2 (a^twob - 2z⁴Y^two). [to⁴b^two + 3rd^twob.z⁴Y^two + 2z⁸Y⁴]

Example 3

Factor (1/125) .x⁶  - 27y⁹ it is much simpler than the previous case. First the equivalents of a and b are identified:

a = (1/5) x^two

b = 3y³

Then they are directly substituted in the formula:

(1/125) .x⁶  - 27y⁹ = [(1/5) x^two - 3y³]. [(1/25) x⁴ + (3/5) x^twoY³ + 9y⁶]

Exercise resolved

The difference of cubes has, as we have said, a variety of applications in Algebra. Let's see some:

Exercise 1

Solve the following equations:

a) x⁵ - 125 x^two = 0

b) 64 - 729 x³ = 0

Solution to

First the equation is factored in this way:

x^two (x³ - 125) = 0

Since 125 is a perfect cube, the parentheses are written as a difference of cubes:

x^two . (x³ - 5³) = 0

The first solution is x = 0, but we find more if we do x³ - 5³ = 0, then:

x³ = 5³ → x = 5

Solution b

The left side of the equation is rewritten as 64 - 729 x³ = 4³ - (9x)³. Therefore:

4³ - (9x)³ = 0

Since the exponent is the same:

9x = 4 → x = 9/4

Exercise 2

Factor the expression:

(x + y)³ - (x - y)³

Solution

This expression is a difference of cubes, if in the factoring formula we note that:

a = x + y

b = x- y

Then the binomial is constructed first:

a - b = x + y - (x- y) = 2y

And now the trinomial:

to^two + a.b + b^two = (x + y)^two + (x + y) (x-y) + (x-y)^two

Notable products are developed:

(x + y)^two = x^two + 2xy + and^two

(x + y) (x-y) = x^two- Y^two

(x- y)^two = x^two - 2xy + and^two

Next you have to substitute and reduce like terms:

to^two + a.b + b^two = x^two + 2xy + and^two+ x^two- Y^two+ x^two - 2xy + and^two = 3x^two + Y^two

Factoring results in:

(x + y)³ - (x - y)³ = 2y. (3x^two + Y^two)

References

1. Baldor, A. 1974. Algebra. Editorial Cultural Venezolana S.A.
2. CK-12 Foundation. Sum and difference of cubes. Recovered from: ck12.org.
3. Khan Academy. Factoring of differences of cubes. Recovered from: es.khanacademy.org.
4. Math is Fun Advanced. Difference of two cubes. Recovered from: mathsisfun.com
5. UNAM. Factoring a difference of cubes. Recovered from: dcb.fi-c.unam.mx.