Before looking at factoring a sum of two cubes, let’s look at the possible factors.
It turns out that
a
3
+
b
3
can actually be factored as (
a
+
b
)(
a
2
–
ab
+
b
2
). Let’s check these factors by multiplying.
Does (
a
+
b
)(
a
2
–
ab
+
b
2
) =
a
3
+
b
3
?
(
a
)(
a
2
–
ab
+
b
2
) + (
b
)(
a
2
–
ab
+
b
2
)
Apply the distributive property.
(
a
3
–
a
2
b
+
ab
2
) + (
b
)(
a
2

ab
+
b
2
)
Multiply by
a
.
(
a
3
–
a
2
b
+
ab
2
) + (
a
2
b
–
ab
2
+
b
3
)
Multiply by
b
.
a
3
–
a
2
b
+
a
2
b + ab
2
–
ab
2
+
b
3
Rearrange terms in order to combine the like
terms.
a
3
+
b
3
Simplify
Did you see that? Four of the terms cancelled out, leaving us with the (seemingly) simple binomial
a
3
+
b
3
. So, the factors are correct.
You can use this pattern to factor binomials in the form
a
3
+
b
3
, otherwise known as “the sum of cubes.”
The Sum of Cubes
A binomial in the form
a
3
+
b
3
can be factored as (
a
+
b
)(
a
2
–
ab
+
b
2
).
Examples:
The factored form of
x
3
+ 64 is (
x
+ 4)(
x
2
– 4
x
+ 16).
The factored form of 8
x
3
+
y
3
is (2
x
+ y)(4
x
2
– 2
xy
+
y
2
).
Example
Problem
Factor
x
3
+ 8
y
3
.
x
3
+ 8
y
3
Identify that this binomial fits the sum of cubes pattern:
a
3
+
b
3
.
a
=
x
, and
b
= 2
y
(since 2
y
• 2
y
• 2
y
= 8
y
3
).
(
x
+ 2
y
)(
x
2
–
x
(2
y
) + (2
y
)
2
)
Factor the binomial as
(
a
+
b
)(
a
2
–
ab
+
b
2
), substituting
a
=
x
and
b
= 2
y
into the expression.
(
x
+ 2
y
)(
x
2
–
x
(2
y
) + 4
y
2
)
Square (2
y
)
2
= 4
y
2
.
Answer
(
x
+ 2
y
)(
x
2
– 2
xy
+ 4
y
2
)
Multiply −
x
(2
y
) = −2
xy
(writing the coefficient first.
And that’s it. The binomial
x
3
+ 8
y
3
can be factored as (
x
+ 2
y
)(
x
2
– 2xy + 4
y
2
)! Let’s try another one.
You should always look for a common factor before you follow any of the patterns for factoring.
Example
Problem
Factor 16
m
3
+ 54
n
3
.
16
m
3
+ 54
n
3
Factor out the common factor 2.
2(8
m
3
+ 27
n
3
)
8
m
3
and 27
n
3
are cubes, so you can factor 8
m
3
+ 27
n
3
as the sum of two
cubes:
a
= 2
m
, and
b
= 3
n
.
2
(2
m
+ 3
n
)[(2
m
)
2
– (2
m
)(3
n
) + (3
n
)
2
]
Factor the binomial 8
m
3
+ 27
n
3
substituting
a
= 2
m
and
b
= 3
n
into the
expression
(
a
+
b
)(
a
2
–
ab
+
b
2
)
.
2(2
m
+ 3
n
)[
4
m
2
– (2
m
)(3
n
) +
9
n
2
]
Square: (2
m
)
2
= 4
m
2
and (3
n
)
2
= 9
n
2
.
Answer
2(2
m
+ 3
n
)(4
m
2
– 6
mn
+ 9
n
2
)
Multiply −(2
m
)(3
n
) = −6
mn.
Factor 125
x
3
+ 64.
(5
x
+ 4)(25
x
2
– 20
x
+ 16)
Correct. 5
x
is the cube root of 125
x
3
, and 4 is the cube root of 64. Substituting these values for
a
and
b
, you find (5
x
+ 4)(25
x
2
– 20
x
+ 16).
Difference of Cubes
Having seen how binomials in the form
a
3
+
b
3
can be factored, it should not come as a surprise that binomials in the form
a
3
–
b
3
can be
factored in a similar way.
The Difference of Cubes
A binomial in the form
a
3
–
b
3
can be factored as (
a
–
b
)(
a
2
+
ab
+
b
2
).
Examples:
The factored form of
x
3
– 64 is (
x
– 4)(
x
2
+ 4
x
+ 16).
The factored form of 27
x
3
– 8
y
3
is (3
x
– 2
y
)(9
x
2
+ 6
xy
+ 4
y
2
).
Notice that the basic construction of the factorization is the same as it is for the sum of cubes; the difference is in the + and – signs. Take a
moment to compare the factored form of
a
3
+
b
3
with the factored form of
a
3
–
b
3
.
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 Spring '12
 george
 Quadratic equation, Elementary algebra, Negative and nonnegative numbers