Solve x5-1=0 (that's x to the 5th power)?

Yes, 1 is the only real root of this equation.
But no one has found the other 4 complex roots as yet!
The LHS can be written as
(x-1)(x^4 + x^3 + x^2 + x + 1) = 0.
So x = 1.
To solve the second factor,
divide by x^2 to get
x^2 + x + 1 + 1/x + 1/x^2 = 0. (*)
We will make a substitution to get a quadratic, then
solve back to get x.
Let t = x + 1/x
Then t^2 = x^2 + 2 + 1/x^2
Rewriting (*) in terms of t, we get
t^2 + t -1 = 0.
There's our quadratic.
Its roots are
t = (-1 + √5)/2 and t = (-1 -√5)/2.
Now to find x, we have,
x + 1/x = t
x^2 - t x + 1 = 0.
x = (t + √t^2-4)/2
x = (t - √(t^2-4)/2.
Plugging in the values of t gives, after some tedious calculations,
x = 1/4[-1 + √5 + √(-10 -2√5)]
x = 1/4[-1 + √5 - √(-10 -2√5)]
x = 1/4[-1 - √5 + √(-10 -2√5)]
x = 1/4[-1 - √5 - √(-10 -2√5)]

x^5 - 1 = 0
x^5 = 1
x = ⁵√1
x = ⁵√(1^5)
x = 1

It is typed as X^5 - 1 = 0
Solution:
X^5 - 1 = 0
X^5 = 1

Taking 5th root of both sides of the above equation:
X = Fifth root of 5
X = 1

The answer X = 1

How To Solve X 5

x^5 - 1 = 0
x^5 = 1
x^5 = 1^5
x = 1

=(x-1)(x^4+x^3+x^2+x+1)
x=1

X^5-1=0
X^5=(0+1)
X=1^-5
X=1

x= -1, 1 and +/-1i

x=1

1x1x1x1x1=1
1-1=0

hope i helped)

x^5=0+1
x^5=1
x=(square root 5)1
x=1