Download 221 8–7 Sum or Difference of Two Cubes Sum of Two Cubes

Section 8–7
8–7
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221
Sum or Difference of Two Cubes
Sum or Difference of Two Cubes
Definition
An expression such as
x3 27
is called the sum of two cubes (x3 and 33). In general, when we multiply the binomial (a b)
and the trinomial (a2 ab b2), we obtain
(a b)(a2 ab b2) a3 a2b ab2 a2b ab2 b3
a3 b3
All but the cubed terms drop out, leaving the sum of two cubes.
Sum of Two
Cubes
a3 b3 (a b)(a2 ab b2)
42
Difference of
Two Cubes
a3 b3 (a b)(a2 ab b2)
43
When we recognize that an expression is the sum (or difference) of two cubes, we can write
the factors immediately.
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Example 39: Factor x3 27.
Solution: This expression is the sum of two cubes, x3 33. Substituting into Eq. 42, with a x
and b 3, yields
x3 27 x3 33 (x 3)(x2 3x 9)
same
sign
opposite
sign
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always ◆◆◆
Example 40: Factor 27x3 8y3.
Solution: This expression is the difference of two cubes, (3x)3 (2y)3. Factoring gives us
27x3 8y3 (3x)3 (2y)3 (3x 2y)(9x2 6xy 4y2)
same
sign
always opposite
sign
Common
Error
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The middle term of the trinomials in Eqs. 42 and 43 is often
mistaken as 2ab.
a3 b3 (a b)(a2 2ab b2)
no!
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Chapter 8
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Factors and Factoring
When the powers are multiples of 3, we may be able to factor the expression as the sum or
difference of two cubes.
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Example 41:
a6 b9 (a2)3 (b3)3
Factoring, we obtain
a6 b9 (a2 b3)(a4 a2b3 b6)
Exercise 7
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Sum or Difference of Two Cubes
Factor completely.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
64 x3
1 64y3
2a3 16
a3 27
x3 1
x3 64
x3 1
a3 343
a3 64
x3 343
x3 125
64a3 27
216 8a3
343 27y3
343 64x3
More Difficult Types
Factor completely.
16. 27x9 512
17. y9 64x3
18. 64a12 x15
19. 27x15 8a6
20. 8x6p 8a6
21. 8a6x 125b3x
22. 64x12a 27y15a
23. 64x3n y9n
24. x3y3 z3
Applications
25. The volume of a hollow spherical shell having an inside radius r1 and an outside radius r2
is 43 r23 43 r13. Factor this expression completely.
26. A cistern is in the shape of a hollow cube whose inside dimension is s and whose
outside dimension is S. If it is made of concrete of density d, its weight is dS3 ds3. Factor
completely.
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Review Problems
Case Study Discussion – Checking Your Factoring
In this case, all the factoring is correct, yet we still know that something is very wrong. In
the fourth line, we have a simple cross multiplication to simplify the expression; however,
the denominator is x y. Since x y, x y has to be zero, and in our mathematical
system, dividing by zero is undefined. What undefined means is that we cannot write a
mathematical definition for dividing by zero.
Sometimes we are so busy factoring we forget a very basic rule. By using the step:
(x y) (x y) x(x y)
we are really dividing BOTH sides by “(x y)”.”
(x y)
(x y)
This leads us to the problem with the next step: if x y, then the denominator here must
be zero.
It would really look something like this:
(x y)(x y) x(x y)
, and we know that anything divided by 0 (zero) is undefined.
(“0”)
(“0”)
So, now this means that the next line, x y = x can’t be true.
This is why you have to be careful when factoring mathematical expressions and
equations.
This particular equation is unfactorable.
In this particular equation where the result is 2 1, we knew there was a problem; in
a different equation, the result might not have been that obvious. This is why in complex
mathematics there is a lot of checking, to ensure that this type of mistake does not happen.
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CHAPTER 8 REVIEW PROBLEMS
Factor completely.
1. x2 2x 15
2. 2a2 3a 2
3. x6 y4
4. 2ax2 8ax 8a
5. 2x2 3x 2
6. a2 ab 6b2
7. (a b)3 (c d)3
y3
8. 8x3 27
x2 x
9. y
y
10. 2ax2y2 18a
11. xy 2y 5x 10
12. 3a2 2a 8
13. x bx y by
2a2 8b2
14. 12
27
15.
16.
17.
18.
(y 2)2 z2
2x2 20ax 50a2
x2 7x 12
4a2 (3a 1)2
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