What are the 8 solutions forx in x^8-1=0?

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

x=1
x=-1
x=i
x=-i
are 4 obvious solutions


x^4 + 1 = 0
x² = ±i
x = a+bi
x² = (a²-b²)+2abi
a²-b² = 0
|a|=|b|
2ab = 1
ab = 1/2
|a|=|b| = 1/√2
a = ±1/√2
b = ±1/√2
x = ±(1/√2)(1±i)

8 solutions:
1
-1
i
-i
(1/√2)(1+i)
(1/√2)(1-i)
(-1/√2)(1+i)
(-1/√2)(1-i)

Alternatively you could use de Moivre's theorem to reach the same answers

Others have spelled out details for the solutions, so let me add something different, namely the underlying geometry of this problem, which is quite interesting.

Some of the solutions are complex numbers, so the natural place for this problem is the complex plane. The 8 complex numbers (remember, real numbers are complex too) that solve this problem are called the "8th roots of unity", because each when raised to the 8th power equals 1.

Roots of unity always lie on the unit circle in the complex plane. The 8th roots of unity are the vertices of a regular octagon, whose circumradius is 1 and with one vertex the number 1 where the unit circle crosses the x-axis.

This means, that the 8th roots of unity are the numbers cos(k*2pi/8) + i sin(k*2pi/8) where k=0,1,2,3,4,5,6,7. So those are the 8 solutions of your equation. 2pi/8 is just 45 degrees, so you need to find the sine and cosine of multiples of 45 degrees, as radicals involving sqrt(2) to complete the solution.

This whole problem generalizes, when viewed in this way to the n-th roots of unity, which are the n complex numbers satisfying the equation x^n - 1 = 0. It also, if you pursue it further, provides a nice introduction to multiplication of complex numbers viewed geometrically, and to certain areas of abstract algebra and number theory, such and finite fields and Gauss sums. But, that is enough for now.

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

x^4 + 1 = 0
x^4 = -1
x = ⁴√(-1)
x = ⁴√1i

x^2 + 1 = 0
x^2 = -1
x = ±√(-1)
x = ±√1i

x + 1 = 0
x = -1

x - 1 = 0
x = 1

∴ x = ⁴√1i , ±√1i , ±1

x^8 -1 = 0, so one immediate solution is x = 1

To find others , decompose the expression into its factors:

[(x^4)^2 - 1][(x^4)^2+ 1]
[(X^2)^2 -1][(X^2)^2 +1][(x^4)^2+ 1]
[x^2 - 1][x^2 + 1][(X^2)^2 +1][(x^4)^2+ 1].........I'm sure you can figure out the rest of it from here