Factoring a Sum or Difference of Two Cubes
This product will lead us to another factoring pattern.
(a + b)(a^{2}  ab + b^{2})
To find the product, multiply each term in (a + b)
by each term in (a^{2}  ab + b^{2}).
To find the product, multiply each term in (a + b)
by each term in (a^{2}  ab + b^{2}).
= a Â· a^{2}  a
Â· ab + a
Â· b^{2} + b Â· a^{2}
 b Â· ab + b
Â· b^{2}
= a^{3}  a^{2}b + ab^{2} + a^{2}b  ab^{2}
+ b^{3}
Combine like terms.
= a^{3} + b^{3}
The four middle terms add to zero.
The result is a^{3} + b^{3}, the sum of two cubes.
Note the structure of the two terms:
â€¢ The first term, a^{3}, is a perfect cube.
â€¢ The last term, b^{3}, is a perfect cube.
â€¢ The terms are added.
When we recognize this pattern, we can immediately factor the sum of two
cubes as follows:
a^{3} + b^{3} = (a + b)(a^{2}  ab + b^{2})
A similar factoring pattern holds for a^{3}  b^{3}, the difference of two cubes.
Note:
You may want to memorize a few
perfect cubes.
1^{3} = 1
2^{3} = 8
3^{3} = 27
4^{3} = 64
5^{3} = 125
To check if a number is a perfect cube, use
the key on your calculator to see if the
cube root is an integer.
Pattern â€”
To Factor the Sum or Difference of Two Cubes
a^{3} + b^{3} = (a + b)(a^{2}  ab + b^{2})
a^{3}  b^{3} = (a  b)(a^{2} + ab + b^{2})
Example 1
Factor: y^{3} + 64
Solution
Step 1 Decide if the given polynomial fits a pattern.
The first term, y^{3}, is a perfect cube, (y)^{3}.
The last term, 64, is a perfect cube, (4)^{3}.
The terms are added.
Therefore, y^{3} + 64 is a sum of two cubes.
Step 2 Identify a and b. Then substitute in the pattern and simplify.
In the factoring pattern for
a sum of two cubes, substitute
y for a and 4 for b.

a^{3} + b^{3} 
= (a + b)(a^{2}  ab + b^{2}) 
Simplify. 
(y)^{3} + (4)^{3} 
= (y + 4)(y^{2}  y Â·
4 + 4^{2})
= (y + 4)(y^{2}  4y + 16) 
The result is:
y^{3} + 64 = (y + 4)(y^{2}  4y + 16).
You can multiply to check the factorization. We leave the check to you.
Example 2
Factor: 8w^{3}  1
Solution
Step 1 Decide if the given polynomial fits a pattern.
The first term, 8w^{3}, is a perfect cube, (2w)^{3}.
The last term, 1, is a perfect cube, (1)^{3}.
The terms are subtracted.
Therefore, 8w^{3}  1 is a difference of two cubes.
Step 2 Identify a and b. Then substitute in the pattern and simplify.
In the factoring pattern for a difference of two cubes,
substitute 2w for a and 1 for b.
Simplify. 
a^{3}  b^{3}
(2w)^{3}  (1)^{3} 
= (a  b)(a^{2} + ab + b^{2})
= (2w  1)[(2w)^{2} + 2w Â· 1
+ (1)^{2}]
= (2w  1)(4w^{2} + 2w + 1) 
The result is:
8w^{3}  1 = (2w  1)(4w^{2} + 2w + 1).
You can multiply to check the factorization. We leave the check to you.
