Zeros of a Difference of Powers

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Key Ideas Easy Medium Hard

x³ − a³ = (x − a)(x² + ax + a²)

x⁶ − a⁶ = (x³ − a³)(x³ + a³)

What to Remember

A cube

x³ − a³ = (x − a)(x² + ax + a²)

x³ + a³ = (x + a)(x² − ax + a²)

Set each factor equal to 0. The quadratic needs the quadratic formula.

A sixth power

x⁶ − a⁶ = (x³ − a³)(x³ + a³). Solve both cubes.

Level 1

Easy

One cube.

Example

Find all zeros of x³ + 8 = 0

x³ + 8 = 0

x³ + 8 = (x)³ + (2)³

a³ + b³ = (a + b)(a² − ab + b²)

a = x

b = 2

(x + 2)(x² − 2x + 4) = 0

x + 2 = 0

x = −2

x² − 2x + 4 = 0

x = −b ± square root of b² − 4ac2a

x = 2 ± square root of 2² − 4(1)(4)2

b² − 4ac = −12

A negative under the square root gives an imaginary number.

square root of −12 = i square root of 12

i square root of 12 = 2isquare root of 3

x = 2 ± 2isquare root of 32

Divide the numerator and the denominator by 2.

x = 1 + isquare root of 3

x = 1 − isquare root of 3

Answer: x = −2
x = 1 + isquare root of 3 or x = 1 − isquare root of 3

Find all zeros of x³ + 64 = 0

Show Answer

x³ + 64 = 0

x³ + 64 = (x)³ + (4)³

a³ + b³ = (a + b)(a² − ab + b²)

a = x

b = 4

(x + 4)(x² − 4x + 16) = 0

x + 4 = 0

x = −4

x² − 4x + 16 = 0

x = −b ± square root of b² − 4ac2a

x = 4 ± square root of 4² − 4(1)(16)2

b² − 4ac = −48

A negative under the square root gives an imaginary number.

square root of −48 = i square root of 48

i square root of 48 = 4isquare root of 3

x = 4 ± 4isquare root of 32

Divide the numerator and the denominator by 2.

x = 2 + 2isquare root of 3

x = 2 − 2isquare root of 3

Answer: x = −4
x = 2 + 2isquare root of 3 or x = 2 − 2isquare root of 3

Find all zeros of x³ + 125 = 0

Show Answer

x³ + 125 = 0

x³ + 125 = (x)³ + (5)³

a³ + b³ = (a + b)(a² − ab + b²)

a = x

b = 5

(x + 5)(x² − 5x + 25) = 0

x + 5 = 0

x = −5

x² − 5x + 25 = 0

x = −b ± square root of b² − 4ac2a

x = 5 ± square root of 5² − 4(1)(25)2

b² − 4ac = −75

A negative under the square root gives an imaginary number.

square root of −75 = i square root of 75

i square root of 75 = 5isquare root of 3

x = 5 ± 5isquare root of 32

x = 5 + 5isquare root of 32

x = 5 − 5isquare root of 32

Answer: x = −5
x = 5 + 5isquare root of 32 or x = 5 − 5isquare root of 32

Find all zeros of x³ + 216 = 0

Show Answer

x³ + 216 = 0

x³ + 216 = (x)³ + (6)³

a³ + b³ = (a + b)(a² − ab + b²)

a = x

b = 6

(x + 6)(x² − 6x + 36) = 0

x + 6 = 0

x = −6

x² − 6x + 36 = 0

x = −b ± square root of b² − 4ac2a

x = 6 ± square root of 6² − 4(1)(36)2

b² − 4ac = −108

A negative under the square root gives an imaginary number.

square root of −108 = i square root of 108

i square root of 108 = 6isquare root of 3

x = 6 ± 6isquare root of 32

Divide the numerator and the denominator by 2.

x = 3 + 3isquare root of 3

x = 3 − 3isquare root of 3

Answer: x = −6
x = 3 + 3isquare root of 3 or x = 3 − 3isquare root of 3

Find all zeros of x³ + 1000 = 0

Show Answer

x³ + 1000 = 0

x³ + 1000 = (x)³ + (10)³

a³ + b³ = (a + b)(a² − ab + b²)

a = x

b = 10

(x + 10)(x² − 10x + 100) = 0

x + 10 = 0

x = −10

x² − 10x + 100 = 0

x = −b ± square root of b² − 4ac2a

x = 10 ± square root of 10² − 4(1)(100)2

b² − 4ac = −300

A negative under the square root gives an imaginary number.

square root of −300 = i square root of 300

i square root of 300 = 10isquare root of 3

x = 10 ± 10isquare root of 32

Divide the numerator and the denominator by 2.

x = 5 + 5isquare root of 3

x = 5 − 5isquare root of 3

Answer: x = −10
x = 5 + 5isquare root of 3 or x = 5 − 5isquare root of 3

Find all zeros of x³ − 343 = 0

Show Answer

x³ − 343 = 0

x³ − 343 = (x)³ − (7)³

a³ − b³ = (a − b)(a² + ab + b²)

a = x

b = 7

(x − 7)(x² + 7x + 49) = 0

x − 7 = 0

x = 7

x² + 7x + 49 = 0

x = −b ± square root of b² − 4ac2a

x = −7 ± square root of 7² − 4(1)(49)2

b² − 4ac = −147

A negative under the square root gives an imaginary number.

square root of −147 = i square root of 147

i square root of 147 = 7isquare root of 3

x = −7 ± 7isquare root of 32

x = −7 + 7isquare root of 32

x = −7 − 7isquare root of 32

Answer: x = 7
x = −7 + 7isquare root of 32 or x = −7 − 7isquare root of 32

Level 2

Medium

Solve the cube, then use it for the sixth power.

Example

Find all zeros of x³ − 8 = 0 Hence find all zeros of x⁶ − 64 = 0

Factor the sixth power as a difference of squares in x³.

x⁶ − 64 = (x³ − 8)(x³ + 8)

Set each factor equal to 0.

x³ − 8 = 0

x³ − 8 = (x)³ − (2)³

a³ − b³ = (a − b)(a² + ab + b²)

a = x

b = 2

(x − 2)(x² + 2x + 4) = 0

x − 2 = 0

x = 2

x² + 2x + 4 = 0

x = −b ± square root of b² − 4ac2a

x = −2 ± square root of 2² − 4(1)(4)2

b² − 4ac = −12

A negative under the square root gives an imaginary number.

square root of −12 = i square root of 12

i square root of 12 = 2isquare root of 3

x = −2 ± 2isquare root of 32

Divide the numerator and the denominator by 2.

x = −1 + isquare root of 3

x = −1 − isquare root of 3

x³ + 8 = 0

x³ + 8 = (x)³ + (2)³

a³ + b³ = (a + b)(a² − ab + b²)

a = x

b = 2

(x + 2)(x² − 2x + 4) = 0

x + 2 = 0

x = −2

x² − 2x + 4 = 0

x = −b ± square root of b² − 4ac2a

x = 2 ± square root of 2² − 4(1)(4)2

b² − 4ac = −12

A negative under the square root gives an imaginary number.

square root of −12 = i square root of 12

i square root of 12 = 2isquare root of 3

x = 2 ± 2isquare root of 32

Divide the numerator and the denominator by 2.

x = 1 + isquare root of 3

x = 1 − isquare root of 3

Answer: x = 2
x = −1 + isquare root of 3 or x = −1 − isquare root of 3
x = −2
x = 1 + isquare root of 3 or x = 1 − isquare root of 3

Find all zeros of x³ − 125 = 0 Hence find all zeros of x⁶ − 15625 = 0

Show Answer

Factor the sixth power as a difference of squares in x³.

x⁶ − 15625 = (x³ − 125)(x³ + 125)

Set each factor equal to 0.

x³ − 125 = 0

x³ − 125 = (x)³ − (5)³

a³ − b³ = (a − b)(a² + ab + b²)

a = x

b = 5

(x − 5)(x² + 5x + 25) = 0

x − 5 = 0

x = 5

x² + 5x + 25 = 0

x = −b ± square root of b² − 4ac2a

x = −5 ± square root of 5² − 4(1)(25)2

b² − 4ac = −75

A negative under the square root gives an imaginary number.

square root of −75 = i square root of 75

i square root of 75 = 5isquare root of 3

x = −5 ± 5isquare root of 32

x = −5 + 5isquare root of 32

x = −5 − 5isquare root of 32

x³ + 125 = 0

x³ + 125 = (x)³ + (5)³

a³ + b³ = (a + b)(a² − ab + b²)

a = x

b = 5

(x + 5)(x² − 5x + 25) = 0

x + 5 = 0

x = −5

x² − 5x + 25 = 0

x = −b ± square root of b² − 4ac2a

x = 5 ± square root of 5² − 4(1)(25)2

b² − 4ac = −75

A negative under the square root gives an imaginary number.

square root of −75 = i square root of 75

i square root of 75 = 5isquare root of 3

x = 5 ± 5isquare root of 32

x = 5 + 5isquare root of 32

x = 5 − 5isquare root of 32

Answer: x = 5
x = −5 + 5isquare root of 32 or x = −5 − 5isquare root of 32
x = −5
x = 5 + 5isquare root of 32 or x = 5 − 5isquare root of 32

Find all zeros of x³ − 216 = 0 Hence find all zeros of x⁶ − 46656 = 0

Show Answer

Factor the sixth power as a difference of squares in x³.

x⁶ − 46656 = (x³ − 216)(x³ + 216)

Set each factor equal to 0.

x³ − 216 = 0

x³ − 216 = (x)³ − (6)³

a³ − b³ = (a − b)(a² + ab + b²)

a = x

b = 6

(x − 6)(x² + 6x + 36) = 0

x − 6 = 0

x = 6

x² + 6x + 36 = 0

x = −b ± square root of b² − 4ac2a

x = −6 ± square root of 6² − 4(1)(36)2

b² − 4ac = −108

A negative under the square root gives an imaginary number.

square root of −108 = i square root of 108

i square root of 108 = 6isquare root of 3

x = −6 ± 6isquare root of 32

Divide the numerator and the denominator by 2.

x = −3 + 3isquare root of 3

x = −3 − 3isquare root of 3

x³ + 216 = 0

x³ + 216 = (x)³ + (6)³

a³ + b³ = (a + b)(a² − ab + b²)

a = x

b = 6

(x + 6)(x² − 6x + 36) = 0

x + 6 = 0

x = −6

x² − 6x + 36 = 0

x = −b ± square root of b² − 4ac2a

x = 6 ± square root of 6² − 4(1)(36)2

b² − 4ac = −108

A negative under the square root gives an imaginary number.

square root of −108 = i square root of 108

i square root of 108 = 6isquare root of 3

x = 6 ± 6isquare root of 32

Divide the numerator and the denominator by 2.

x = 3 + 3isquare root of 3

x = 3 − 3isquare root of 3

Answer: x = 6
x = −3 + 3isquare root of 3 or x = −3 − 3isquare root of 3
x = −6
x = 3 + 3isquare root of 3 or x = 3 − 3isquare root of 3

Find all zeros of x³ − 1000 = 0 Hence find all zeros of x⁶ − 1000000 = 0

Show Answer

Factor the sixth power as a difference of squares in x³.

x⁶ − 1000000 = (x³ − 1000)(x³ + 1000)

Set each factor equal to 0.

x³ − 1000 = 0

x³ − 1000 = (x)³ − (10)³

a³ − b³ = (a − b)(a² + ab + b²)

a = x

b = 10

(x − 10)(x² + 10x + 100) = 0

x − 10 = 0

x = 10

x² + 10x + 100 = 0

x = −b ± square root of b² − 4ac2a

x = −10 ± square root of 10² − 4(1)(100)2

b² − 4ac = −300

A negative under the square root gives an imaginary number.

square root of −300 = i square root of 300

i square root of 300 = 10isquare root of 3

x = −10 ± 10isquare root of 32

Divide the numerator and the denominator by 2.

x = −5 + 5isquare root of 3

x = −5 − 5isquare root of 3

x³ + 1000 = 0

x³ + 1000 = (x)³ + (10)³

a³ + b³ = (a + b)(a² − ab + b²)

a = x

b = 10

(x + 10)(x² − 10x + 100) = 0

x + 10 = 0

x = −10

x² − 10x + 100 = 0

x = −b ± square root of b² − 4ac2a

x = 10 ± square root of 10² − 4(1)(100)2

b² − 4ac = −300

A negative under the square root gives an imaginary number.

square root of −300 = i square root of 300

i square root of 300 = 10isquare root of 3

x = 10 ± 10isquare root of 32

Divide the numerator and the denominator by 2.

x = 5 + 5isquare root of 3

x = 5 − 5isquare root of 3

Answer: x = 10
x = −5 + 5isquare root of 3 or x = −5 − 5isquare root of 3
x = −10
x = 5 + 5isquare root of 3 or x = 5 − 5isquare root of 3

Find all zeros of x³ − 512 = 0 Hence find all zeros of x⁶ − 262144 = 0

Show Answer

Factor the sixth power as a difference of squares in x³.

x⁶ − 262144 = (x³ − 512)(x³ + 512)

Set each factor equal to 0.

x³ − 512 = 0

x³ − 512 = (x)³ − (8)³

a³ − b³ = (a − b)(a² + ab + b²)

a = x

b = 8

(x − 8)(x² + 8x + 64) = 0

x − 8 = 0

x = 8

x² + 8x + 64 = 0

x = −b ± square root of b² − 4ac2a

x = −8 ± square root of 8² − 4(1)(64)2

b² − 4ac = −192

A negative under the square root gives an imaginary number.

square root of −192 = i square root of 192

i square root of 192 = 8isquare root of 3

x = −8 ± 8isquare root of 32

Divide the numerator and the denominator by 2.

x = −4 + 4isquare root of 3

x = −4 − 4isquare root of 3

x³ + 512 = 0

x³ + 512 = (x)³ + (8)³

a³ + b³ = (a + b)(a² − ab + b²)

a = x

b = 8

(x + 8)(x² − 8x + 64) = 0

x + 8 = 0

x = −8

x² − 8x + 64 = 0

x = −b ± square root of b² − 4ac2a

x = 8 ± square root of 8² − 4(1)(64)2

b² − 4ac = −192

A negative under the square root gives an imaginary number.

square root of −192 = i square root of 192

i square root of 192 = 8isquare root of 3

x = 8 ± 8isquare root of 32

Divide the numerator and the denominator by 2.

x = 4 + 4isquare root of 3

x = 4 − 4isquare root of 3

Answer: x = 8
x = −4 + 4isquare root of 3 or x = −4 − 4isquare root of 3
x = −8
x = 4 + 4isquare root of 3 or x = 4 − 4isquare root of 3

Find all zeros of x³ − 343 = 0 Hence find all zeros of x⁶ − 117649 = 0

Show Answer

Factor the sixth power as a difference of squares in x³.

x⁶ − 117649 = (x³ − 343)(x³ + 343)

Set each factor equal to 0.

x³ − 343 = 0

x³ − 343 = (x)³ − (7)³

a³ − b³ = (a − b)(a² + ab + b²)

a = x

b = 7

(x − 7)(x² + 7x + 49) = 0

x − 7 = 0

x = 7

x² + 7x + 49 = 0

x = −b ± square root of b² − 4ac2a

x = −7 ± square root of 7² − 4(1)(49)2

b² − 4ac = −147

A negative under the square root gives an imaginary number.

square root of −147 = i square root of 147

i square root of 147 = 7isquare root of 3

x = −7 ± 7isquare root of 32

x = −7 + 7isquare root of 32

x = −7 − 7isquare root of 32

x³ + 343 = 0

x³ + 343 = (x)³ + (7)³

a³ + b³ = (a + b)(a² − ab + b²)

a = x

b = 7

(x + 7)(x² − 7x + 49) = 0

x + 7 = 0

x = −7

x² − 7x + 49 = 0

x = −b ± square root of b² − 4ac2a

x = 7 ± square root of 7² − 4(1)(49)2

b² − 4ac = −147

A negative under the square root gives an imaginary number.

square root of −147 = i square root of 147

i square root of 147 = 7isquare root of 3

x = 7 ± 7isquare root of 32

x = 7 + 7isquare root of 32

x = 7 − 7isquare root of 32

Answer: x = 7
x = −7 + 7isquare root of 32 or x = −7 − 7isquare root of 32
x = −7
x = 7 + 7isquare root of 32 or x = 7 − 7isquare root of 32

Level 3

Hard

A coefficient on x, or a sixth power with no hint.

Example

Find all zeros of 8x³ − 1 = 0

8x³ − 1 = 0

8x³ − 1 = (2x)³ − (1)³

a³ − b³ = (a − b)(a² + ab + b²)

a = 2x

b = 1

(2x − 1)(4x² + 2x + 1) = 0

2x − 1 = 0

2x = 1

x = 12

4x² + 2x + 1 = 0

x = −b ± square root of b² − 4ac2a

x = −2 ± square root of 2² − 4(4)(1)8

b² − 4ac = −12

A negative under the square root gives an imaginary number.

square root of −12 = i square root of 12

i square root of 12 = 2isquare root of 3

x = −2 ± 2isquare root of 38

Divide the numerator and the denominator by 2.

x = −1 + isquare root of 34

x = −1 − isquare root of 34

Answer: x = 12
x = −1 + isquare root of 34 or x = −1 − isquare root of 34

Find all zeros of 8x³ + 1 = 0

Show Answer

8x³ + 1 = 0

8x³ + 1 = (2x)³ + (1)³

a³ + b³ = (a + b)(a² − ab + b²)

a = 2x

b = 1

(2x + 1)(4x² − 2x + 1) = 0

2x + 1 = 0

2x = −1

x = −12

4x² − 2x + 1 = 0

x = −b ± square root of b² − 4ac2a

x = 2 ± square root of 2² − 4(4)(1)8

b² − 4ac = −12

A negative under the square root gives an imaginary number.

square root of −12 = i square root of 12

i square root of 12 = 2isquare root of 3

x = 2 ± 2isquare root of 38

Divide the numerator and the denominator by 2.

x = 1 + isquare root of 34

x = 1 − isquare root of 34

Answer: x = −12
x = 1 + isquare root of 34 or x = 1 − isquare root of 34

Find all zeros of 27x³ − 8 = 0

Show Answer

27x³ − 8 = 0

27x³ − 8 = (3x)³ − (2)³

a³ − b³ = (a − b)(a² + ab + b²)

a = 3x

b = 2

(3x − 2)(9x² + 6x + 4) = 0

3x − 2 = 0

3x = 2

x = 23

9x² + 6x + 4 = 0

x = −b ± square root of b² − 4ac2a

x = −6 ± square root of 6² − 4(9)(4)18

b² − 4ac = −108

A negative under the square root gives an imaginary number.

square root of −108 = i square root of 108

i square root of 108 = 6isquare root of 3

x = −6 ± 6isquare root of 318

Divide the numerator and the denominator by 6.

x = −1 + isquare root of 33

x = −1 − isquare root of 33

Answer: x = 23
x = −1 + isquare root of 33 or x = −1 − isquare root of 33

Find all zeros of 27x³ + 8 = 0

Show Answer

27x³ + 8 = 0

27x³ + 8 = (3x)³ + (2)³

a³ + b³ = (a + b)(a² − ab + b²)

a = 3x

b = 2

(3x + 2)(9x² − 6x + 4) = 0

3x + 2 = 0

3x = −2

x = −23

9x² − 6x + 4 = 0

x = −b ± square root of b² − 4ac2a

x = 6 ± square root of 6² − 4(9)(4)18

b² − 4ac = −108

A negative under the square root gives an imaginary number.

square root of −108 = i square root of 108

i square root of 108 = 6isquare root of 3

x = 6 ± 6isquare root of 318

Divide the numerator and the denominator by 6.

x = 1 + isquare root of 33

x = 1 − isquare root of 33

Answer: x = −23
x = 1 + isquare root of 33 or x = 1 − isquare root of 33

Find all zeros of 64x³ − 1 = 0

Show Answer

64x³ − 1 = 0

64x³ − 1 = (4x)³ − (1)³

a³ − b³ = (a − b)(a² + ab + b²)

a = 4x

b = 1

(4x − 1)(16x² + 4x + 1) = 0

4x − 1 = 0

4x = 1

x = 14

16x² + 4x + 1 = 0

x = −b ± square root of b² − 4ac2a

x = −4 ± square root of 4² − 4(16)(1)32

b² − 4ac = −48

A negative under the square root gives an imaginary number.

square root of −48 = i square root of 48

i square root of 48 = 4isquare root of 3

x = −4 ± 4isquare root of 332

Divide the numerator and the denominator by 4.

x = −1 + isquare root of 38

x = −1 − isquare root of 38

Answer: x = 14
x = −1 + isquare root of 38 or x = −1 − isquare root of 38

Find all zeros of x⁶ − 4096 = 0

Show Answer

x⁶ − 4096 = (x³ − 64)(x³ + 64)

Set each factor equal to 0.

x³ − 64 = 0

x³ − 64 = (x)³ − (4)³

a³ − b³ = (a − b)(a² + ab + b²)

a = x

b = 4

(x − 4)(x² + 4x + 16) = 0

x − 4 = 0

x = 4

x² + 4x + 16 = 0

x = −b ± square root of b² − 4ac2a

x = −4 ± square root of 4² − 4(1)(16)2

b² − 4ac = −48

A negative under the square root gives an imaginary number.

square root of −48 = i square root of 48

i square root of 48 = 4isquare root of 3

x = −4 ± 4isquare root of 32

Divide the numerator and the denominator by 2.

x = −2 + 2isquare root of 3

x = −2 − 2isquare root of 3

x³ + 64 = 0

x³ + 64 = (x)³ + (4)³

a³ + b³ = (a + b)(a² − ab + b²)

a = x

b = 4

(x + 4)(x² − 4x + 16) = 0

x + 4 = 0

x = −4

x² − 4x + 16 = 0

x = −b ± square root of b² − 4ac2a

x = 4 ± square root of 4² − 4(1)(16)2

b² − 4ac = −48

A negative under the square root gives an imaginary number.

square root of −48 = i square root of 48

i square root of 48 = 4isquare root of 3

x = 4 ± 4isquare root of 32

Divide the numerator and the denominator by 2.

x = 2 + 2isquare root of 3

x = 2 − 2isquare root of 3

Answer: x = 4
x = −2 + 2isquare root of 3 or x = −2 − 2isquare root of 3
x = −4
x = 2 + 2isquare root of 3 or x = 2 − 2isquare root of 3

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