Factoring the Sum and Difference of Cubes

 In mathematics, factoring is one important topic in algebra. Factoring the sum and difference of cubes are the one special case of polynomials in factoring.. Factoring is the process of performing the operation of arithmetic multiplication in reverse. Let us solve some example problems for factoring the sum and difference of cubes.

 

Factoring the Sum and Difference of Cubes:

 

Sum of cubes:

(a3 + b3) = (a + b) (a2 - ab + b2)

Differences of cubes:

(a3 – b3) = (a – b) (a2 + ab + b2)

 

Example Problems for Factoring the Sum and Difference of Cubes:

 

Different example problems for factoring the sum and difference of cubes are,

Problem 1:

Factor the given polynomial expression

8x– 64

Solution:

Given polynomial expression

8x– 64

It is in the form of difference of cubes, because the expression is in the form

(2x)3 – (4)3

Factor the expression

The first term of the expression can be written as 

8x= 2x . (2x)2

The last term of the expression can be written as

64 = 4 . 42

Now we factor the given expression

(a3 – b3) = (a – b) (a2 + ab + b2)

In the given expression,

a = 2x and b = 4

8x– 64 = (2x -4) (4x2 + 2x (4) + 42)

             = (2x – 4) (4x2 + 8x + 16)

Answer:

8x– 64 = (2x – 4) (4x2 + 8x + 16)

Problem  2:

Factor the given polynomial expression

64m– 1000

Solution:

Given polynomial expression

64m– 1000

It is in the form of difference of cubes, because the expression is in the form

(4m)3 – (10)3

Factor the expression

The first term of the expression can be written as 

(4m)= 4m . (4m)2

The last term of the expression can be written as

1000 = 10 . (10)2

Now we factor the given expression

(a3 – b3) = (a – b) (a2 + ab + b2)

In the given expression,

a = 4m and b = 10

64m– 1000 = (4m - 10) (16m2 + 4m (10) + (10)2)

             = (4m – 10) (16m2+ 40m + 100)

Answer:

64m– 1000 = (4m – 10) (16m2 + 40m + 100)

 

Problem 3:

Factor the given polynomial expression

27x+ 8

Solution:

Given polynomial expression

27x+ 8

It is in the form of sum of cubes, because

(3x)3 + (2)3

Factor the terms

The first term of the expression can be written as 

27x= 3x . (3x)2

The last term of the expression can be written as

8 = 2 . (2)2

Now we factor the given expression

(a3 + b3) = (a + b) (a2 - ab + b2)

In the given expression,

a = 3x and b = 2

27x+ 8 = (3x + 2) (9x2 – 3x (2) + (2)2)

              = (3x + 2) (9x2 – 6x + 4)

Answer:

27x+ 8 = (3x + 2) (9x2 – 6x + 4)

Problem 4:

Factor the given polynomial expression

125m+ 216

Solution:

Given polynomial expression

125m+ 216

It is in the form of sum of cubes, because

(5m)3 + (6)3

Factor the terms

The first term of the expression can be written as 

125m= 5m . (5m)2

The last term of the expression can be written as

216 = 6 . 62

Now we factor the given expression

(a3 + b3) = (a + b) (a2 - ab + b2)

In the given expression,

a = 5m and b = 6

125m+ 216 = (5m + 6) ((5m)2 – 5m (6) + 62)

                     = (5m + 6) (25m2 – 30m + 36)

Answer:

125m+ 216 = (5m + 6) (25m2 – 30m + 36)