# math

How (1+4x^-6) equal x^-6 +4?

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1. 1+4x^-6 = x^-6+4
3x^-6 = 3
x^-6 = 1
x^6 = 1.
Take the 6th root of both sides:
X = 1.

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2. x = ±1 since all our powers are even.

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3. 1 + 4 x⁻⁶ = x⁻⁶ + 4

1 + 4 / x⁶ = 1 / x⁶ + 4

Subtract​ 1 from both sides:

4 / x⁶ = 1 / x⁶ + 3

Multiply both sides by x⁶:

4 = 1 + 3 x⁶

Subtract​ 1 from both sides:

3 = 3 x⁶

Divide both sides by 3:

1 = x⁶

Subtract​ 1 from both sides:

0 = x⁶ - 1

x⁶ - 1 = 0

Substitution:

x² = u

x⁶ = u³

x⁶ - 1 = 0

u³ - 1 = 0

u³ - 1³ = 0

Difference of cubes:

a³ - b³ = ( a - b ) ( a² + a ∙ b + b² )

In this case:

u³ - 1³ = ( u - 1 ) ( u² + u ∙ 1 + 1² )

u³ - 1³ = ( u - 1 ) ( u² + u + 1 )

This expression will be equal to zero when both terms are equal to zero.

( u - 1 ) = 0 and u² + u + 1 = 0

u - 1 = 0

u = 1

x² = 1

x = ± √1

x = ±1

The solutions are:

x = - 1 and x = 1

u² + u + 1 = 0

Try to solve that.

The solutions are:

u = - 1 / 2 + √3 i / 2 and u = - 1 / 2 - √3 i / 2

u = ( - 1 + √3 i ) / 2 and u = - ( 1 + √3 i ) / 2

This means that this equation has not only real solutions, there are also complex conjugate solutions.

Substitute back u = x², and solve for x.

x² = - 1 / 2 + √3 i / 2

x² = - ( 1 - √3 i ) / 2 = [ - ( 1 + √3 i ) / 2 ]

x = ± √ [ - ( 1 + √3 i ) / 2 ]²

x = - ( 1 + √3 i ) / 2 and ( 1 + √3 i ) / 2

x² = - ( 1 / 2 + √3 i / 2 )

- ( 1 + √3 i ) / 2 = - ( 1 - √3 i / 2 )²

x = ± √ [ - ( 1 - √3 i ) / 2 ]²

x = - ( 1 - √3 i ) / 2 and - [ - ( 1 - √3 i ) / 2 ]

x = ( - 1 + √3 i ) / 2 and ( 1 - √3 i / ) / 2

Finally the solutions are:

x = - 1 , x = 1 , x = - ( 1 + √3 i ) / 2 , x = ( 1 + √3 i ) / 2 , x = ( - 1 + √3 i ) / 2 , x = ( 1 - √3 i ) / 2

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