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Sum and Difference of Cubes
Lori Jordan
Kate Dirga
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Printed: March 24, 2016
AUTHORS
Lori Jordan
Kate Dirga
www.ck12.org
C HAPTER
Chapter 1. Sum and Difference of Cubes
1
Sum and Difference of
Cubes
Here you’ll learn how to use the sum and difference of cubes formulas for factoring certain types of polynomials.
The volume of a rectangular prism is 2x4 − 128x. What are the lengths of the prism’s sides?
Sum and Difference of Cubes
Previously, you learned how to factor several different types of quadratic equations. Here, we will expand this
knowledge to certain types of polynomials. The first is the sum of cubes. The sum of cubes is what it sounds like,
the sum of two cube numbers or a3 + b3 . We will use an investigation involving volume to find the factorization of
this polynomial.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/177334
Investigation: Sum of Cubes Formula
1. Pictorially, the sum of cubes looks like this:
Or, we can put one on top of the other.
2. Recall that the formula for volume is length × width × depth. Find the volume of the sum of these two cubes.
V = a3 + b3
3. Now, we will find the volume in a different way. Using the second picture above, we will add in imaginary lines
so that these two cubes look like one large prism. Find the volume of this prism.
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V = a × a × (a + b)
= a2 (a + b)
4. Subtract the imaginary portion on top. In the picture, they are prism 1 and prism 2.


V = a2 (a + b) − ab(a − b) + b2 (a − b)
| {z } | {z }
Prism 1
Prism 2
5. Pull out any common factors within the brackets.
V = a2 (a + b) − b(a − b)[a + b]
6. Notice that both terms have a common factor of (a + b). Pull this out, put it in front, and get rid of the brackets.
V = (a + b)(a2 − b(a − b))
7. Simplify what is inside the second set of parenthesis.
V = (a + b)(a2 − ab + b2 )
In the last step, we found that a3 + b3 factors to (a + b)(a2 − ab + b2 ). This is the Sum of Cubes Formula.
Solve the following problems
Factor 8x3 + 27.
First, determine if these are “cube” numbers. A cube number has a cube root. For example, the cube root of 8 is 2
because 23 = 8. 33 = 27, 43 = 64, 53 = 125, and so on.
a3 = 8x3 = (2x)3
a = 2x
2
b3 = 27 = 33
b=3
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Chapter 1. Sum and Difference of Cubes
In the formula, we have:
(a + b)(a2 − ab + b2 ) = (2x + 3)((2x)2 − (2x)(3) + 32 )
= (2x + 3)(4x2 − 6x + 9)
Therefore, 8x3 + 27 = (2x + 3)(4x2 − 6x + 9). The second factored polynomial does not factor any further.
Investigation: Difference of Cubes
1. Pictorially, the difference of cubes looks like this:
Imagine the smaller cube is taken out of the larger cube.
2. Recall that the formula for volume is length × width × depth. Find the volume of the difference of these two
cubes.
V = a3 − b3
3. Now, we will find the volume in a different way. Using the picture here, will add in imaginary lines so that the
shape is split into three prisms. Find the volume of prism 1, prism 2, and prism 3.
Prism 1 : a · a · (a − b)
Prism 2 : a · b · (a − b)
Prism 3 : b · b · (a − b)
4. Add the volumes together to get the volume of the entire shape.
V = a2 (a − b) + ab(a − b) + b2 (a − b)
5. Pull out any common factors and simplify.
V = (a − b)(a2 + ab + b2 )
In the last step, we found that a3 − b3 factors to (a − b)(a2 + ab + b2 ). This is the Difference of Cubes Formula.
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Factor x5 − 125x2 .
First, take out any common factors.
x5 − 125x2 = x2 (x3 − 125)
What is inside the parenthesis is a difference of cubes. Use the formula.
x5 − 125x2 = x2 (x3 − 125)
= x2 (x3 − 53 )
= x2 (x − 5)(x2 + 5x + 25)
Find the real-number solutions of x3 − 8 = 0.
Factor using the difference of cubes.
x3 − 8 = 0
(x − 2)(x2 + 2x + 4) = 0
x=2
In the last step, we set the first factor equal to zero. The second factor, x2 + 2x + 4, will give imaginary solutions.
For both the sum and difference of cubes, this will always happen.
Examples
Example 1
Earlier, you were asked what are the lengths of the prism’s sides.
We need to factor 2x4 − 128x.
First, take out any common factors.
2x4 − 128x = 2x(x3 − 64)
What is inside the parenthesis is a difference of cubes. Use the Difference of Cubes Formula.
2x(x3 − 64)
= 2x(x3 − 43 )
= 2x(x − 4)(x2 + 4x + 16)
Therefore, the side lengths of the rectangular prism are 2x, x + 4, and x2 + 4x + 16.
Factor using the sum or difference of cubes.
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Chapter 1. Sum and Difference of Cubes
Example 2
x3 − 1
Factor using the difference of cubes.
x 3 − 1 = x 3 − 13
= (x − 1)(x2 + x + 1)
Example 3
3x3 + 192
Pull out the 3, then factor using the sum of cubes.
3x3 + 192 = 3(x3 + 64)
= 3(x3 + 43 )
= 3(x + 4)(x2 − 4x + 16)
Example 4
125 − 216x3
Factor using the difference of cubes.
125 − 216x3 = 53 − (6x)3
= (5 − 6x)(52 + (5)(6x) + (6x)2 )
= (5 − 6x)(25 + 30x + 36x2 )
Example 5
Find the real-number solution to 27x3 + 8 = 0.
Factor using the sum of cubes and then solve.
27x3 + 8 = 0
(3x)3 + 23 = 0
(3x + 2)(9x2 − 6x + 4) = 0
x=−
2
3
Review
Factor each polynomial by using the sum or difference of cubes.
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1.
2.
3.
4.
5.
6.
7.
8.
9.
x3 − 27
64 + x3
32x3 − 4
64x3 + 343
512 − 729x3
125x4 + 8x
648x3 + 81
5x6 − 135x3
686x7 − 1024x4
Find the real-number solutions for each equation.
10.
11.
12.
13.
14.
15.
125x3 + 1 = 0
64 − 729x3 = 0
8x4 − 343x = 0
Challenge Find ALL solutions (real and imaginary) for 5x5 + 625x2 = 0.
Challenge Find ALL solutions (real and imaginary) for 686x3 + 2000 = 0.
Real Life Application You have a piece of cardboard that you would like to fold up and make an open (no
top) box out of. The dimensions of the cardboard are 3600 × 4200 . Write a factored equation for the volume of
this box. Find the volume of the box when x = 1, 3, and 5.
Answers for Review Problems
To see the Review answers, open this PDF file and look for section 6.6.
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