1+4x^-6 = x^-6+4
3x^-6 = 3
x^-6 = 1
x^6 = 1.
Take the 6th root of both sides:
X = 1.
How (1+4x^-6) equal x^-6 +4?
3 answers
x = ±1 since all our powers are even.
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
Add 1 to both sides:
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
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
Add 1 to both sides:
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