A Nice Quintic Equation

[@wtspman]

In the first method, where you used the quadratic formula for the 3rd factor, the denominator should have been 2a = 6.

[@shmuelzehavi4940]

The solution to this equation may be written in a compact form as:
x = 1/2 + i (n/6) √3 , {n} = {0 , ± 1 , ± 3}

[@StaR-uw3dc]

Other approach:
Multiply both sides by 2⁶=64, substitute t=2x-1 and get (t+1)⁶=(t-1)⁶.
After expanding, reducing and factoring we get the quintic in t: t(3t²+1)(t²+3)=0 with roots t=0, t=±(√3/3)i, t=±√3i.
Finally x=(t+1)/2 and x=1/2, x=(3±√3i)/6, x=(1±√3i)/2

great. The unique real solution 1/2 is obvious by the symmetry of the problem, so the substitution makes everything flow smoothly

[@Don-Ensley]

problem
x⁶ = (x-1)⁶
(x-1)⁶ - x⁶ = 0
This is quintic because the x⁶ subtracts off of the expansion of (x-1)⁶, so we expect 5 roots.
Factor as a difference of 2 cubes.
[ (x-1)² - x²] [ (x-1)⁴ +(x-1)² x² +x⁴] = 0
Use the zero product property.
First term:
(x-1)² - x² = 0
(x-1-x)(x-1+x) = 0
2x = 1
x = 1/2
Second term:
(x-1)⁴ +(x-1)² x² +x⁴ = 0
(x-1)² [ (x-1)² + x²] + x⁴ = 0
(x²-2x+1)(2x²-2x+1) + x⁴ = 0
3(x²-x)² + 4(x²-x) + 1 = 0
Let
y = x²-x
3y² +4y + 1 = 0
y = [ -4 ± √4 ] / 6
= ( -2 ± 1 ) / 3
= -1, -1/3
x²-x = -1
(x - 1/2) ²-1/4 = -1
x -1/2 = ± i√3 /2
x = 1/2 ± i√3 /2
x²-x = -1/3
(x - 1/2) ²-1/4 = -1/3
(x - 1/2) ² = -1/12
x - 1/2 = ± i √3 / 6
x = 1/2 ± i √3 / 6
answer
x ∈ { 1/2,
(1 - i√3) / 2,
(3 - i√3) / 6,
(3 + i√3) / 6,
(1 + i√3) / 2 }