FACTORING DIFFERENCE OF CUBES

Factor each of the following expressions.

Problem 1 :

x³ - 64

Solution :

= x³ - 64

Decomposing 64 as product of three same terms, we get

64 = 4 ⋅ 4 ⋅ 4

= 4³

= x³ - 4³

= (x - 4)(x² + x(4) + 4²)

= (x - 4)(x² + 4x + 16)

Problem 2 :

8x³ - 27

Solution :

= 8x³ - 27

Decomposing 8 as product of three same terms, we get

8 = 2 ⋅ 2 ⋅ 2

Decomposing 27 as product of three same terms, we get

27 = 3 ⋅ 3 ⋅ 3

= 3³

= 2³x³ - 3³

= (2x)³ - 3³

Here a = 2x and b = 3

= (2x - 3)((2x)² + 2x(3) + 3²)

= (2x - 3)(4x² + 6x + 9)

Problem 3 :

x³ - 125

Solution :

= x³ - 125

Decomposing 125 as product of three same terms, we get

125 = 5 ⋅ 5 ⋅ 5

= 5³

= x³ - 5³

Here a = x and b = 5

= (x - 5)(x² + x(5) + 5²)

= (x - 5)(x² + 5x + 25)

Problem 4 :

64x³ - 125

Solution :

= 64x³ - 125

Decomposing 64 as product of three same terms, we get

64 = 4 ⋅ 4 ⋅ 4

= 4³

Decomposing 125 as product of three same terms, we get

125 = 5 ⋅ 5 ⋅ 5

= 5³

= 4³x³ - 5³

= (4x)³ - 5³

= (4x - 5)((4x)² + 4x(5) + 5²)

= (4x - 5)(16x² + 20x + 25)

Problem 5 :

216x³ - 8

Solution :

= 216x³ - 8

Decomposing 8 as product of three same terms, we get

216 = 6 ⋅ 6 ⋅ 6

Decomposing 8 as product of three same terms, we get

8 = 2 ⋅ 2 ⋅ 2

= 2³

= 6³x³ - 2³

= (6x)³ - 2³

Here a = 6x and b = 2

= (6x - 2)((6x)² + 6x(2) + 2²)

= (6x - 2)(36x² + 12x + 4)

Factoring 2, we get

= 2(3x - 1)(36x² + 12x + 4)

Problem 6 :

(x + 5)³ - (2x + 1)³

Solution :

= (x + 5)³ - (2x + 1)³

Here a = x + 5 and b = 2x + 1

a - b = x + 5 - (2x + 1)

= x + 5 - 2x - 1

= -x + 4

= 4 - x

= (4 - x)[(x + 5)² + (x + 5)(2x + 1) + (2x + 1)²)

Expanding (x + 5)² :

= x² + 2x(5) + 5²

= x² + 10x + 25

Expanding (2x + 1)² :

= (2x)² + 2(2x)(1) + 1²

= 4x² + 4x + 1

Expanding (x + 5)(2x + 1) :

= x² + x + 10x + 5

= x² + 11x + 5

= (4 - x)[x² + 10x + 25 + x² + 11x + 5 + 4x² + 4x + 1]

= (4 - x)[3x² + 25x + 31]

Problem 7 :

- 3x⁴ + 24x

Solution :

= - 3x⁴ + 24x

Factoring -3x

= -3x(x³ - 8)

= -3x(x³ - 2³)

= -3x(x - 2)(x² + x(2) + 2²)

= -3x(x - 2)(x² + 2x + 4)

Problem 8 :

(1/27)x³ - (8/125)

Solution :

= (1/27)x³ - (8/125)

Decomposing 1/27 as product of three same terms, we get

1/27 = 1/3 ⋅ 1/3 ⋅ 1/3

Decomposing 8/125 as product of three same terms, we get

8/125 = 2/5 ⋅ 2/5 ⋅ 2/5

= (1/3)³x³ - (2/5)³

= [(1/3)x]³ - (2/5)³

Here a = 1/3 and b = 2/5

= [(1/3)x - (2/5)] [(x/3)² + (x/3)(2/5) + (2/5)²]

= [(1/3)x - (2/5)] [[(x²/9) + (2x/15) + (4/25))

Problem 9 :

(x - 3)³ - (3x - 2)³

Solution :

= (x - 3)³ - (3x - 2)³

Here a = x - 3 and b = 3x - 2

a - b = x - 3 - (3x - 2)

= x - 3 - 3x + 2

= -2x - 1

= ( -2x - 1)[(x - 3)² + (x - 3)(3x - 2) + (3x - 2)²)

Expanding (x - 3)² :

= x² - 2x(3) + 3²

= x² - 6x + 9

Expanding (3x - 2)² :

= (3x)² - 2(3x)(2) + 2²

= 9x² - 12x + 4

Expanding (x - 3)(3x - 2) :

= 3x² - 2x - 9x + 6

= 3x² - 11x + 6

= (-2x - 1)[x² - 6x + 9 + 9x² - 12x + 4 + 3x² - 11x + 6)

= (-2x - 1)(13x²  - 6x - 12x - 11x + 9 + 4 + 6)

= (-2x - 1)(13x²  - 29x + 19)

Problem 10 :

-432 x⁵ + 128 x²

Solution :

= -432 x⁵ + 128 x²

Greatest common factor of 128 and 432 is 16. Factoring -16x²

= -16x² (27x³ - 8)

= -16x² (3³x³ - 2³)

= -16x² ((3x)³ - 2³)

Here a = 3x and b = 2

= -16x² (3x - 2)((3x)² + 3x(2) + 2²)

= -16x² (3x - 2)(9x² + 6x + 4)