Reciprocal equation

[Guest]

(x²+x+1)⁴+(x²+ 1-x)⁴+x⁸+1=0

[adamsmith]

Given (x2+x+1)4+(x2+ 1-x)4+x8+1=0
(x^4+4x²+1)(3x^4+8x²+3)=0
there for x have 8 complex solution
x1=-i√(2-√3)
x2=i√(2-√3)
x3=-i√(2+√3)
x4=i√(2+√3)
x5=-i√(1/3(4-√7))
x6=i√(1/3(4-√7))
x7=-i√(1/3(4+√7))
x8=i√(1/3(4+√7))

[Mathmaven53]

( x^2 + x + 1)^4 + (x^2 + 1 - x)^4 + x^8 + 1 = 0

Divide by x^4

(x + 1/x + 1)^4 + (x + 1/x - 1)^4 + x^4 + 1/x^4 = 0

Let t = x + 1/x

t^2 = x^2 + 1/x^2 + 2

x^2 + 1/x^2 = t^2 - 2

Square

x^4 + 1/x^4 + 2 = (t^2 - 2)^2

x^4 + 1/x^4 = (t^2 - 2)^2 - 2

= t^4 - 4 t^2 + 2

Substitute into

(x + 1/x + 1)^4 + (x + 1/x - 1)^4 + x^4 + 1/x^4 = 0

(t + 1)^4 + (t -1)^4 + t^4 - 4 t^2 + 2 = 0

Then expanding we get

3 t^4 + 8 t^2 + 4 = 0

Factor

(3 t^2 + 2)(t^2 + 2) = 0

Then set each factor equal to zero and solve for t

For each t found

x + 1/x = t

x^2 + 1 = t x

x^2 - t x + 1 = 0

Then solve this for x