how to solve X^3=1 X=?

For a cubic equation you get 3 answers.

x³ = 1
x³ - 1 = 0
x³-1³=0
(x-1)(x²+x+1) = 0

x - 1=0
x = 1

x²+x+1 = 0
x = (-1±√(1²-4×1×1))/2
= (-1±√-3)/2
= (-1±i√3)/2

The 3 cube roots of 1:
{1, (-1+i√3)/2, (-1-i√3)/2}

Note that [(-1+i√3)/2]² = (-1-i√3)/2
and
[(-1-i√3)/2]² = (-1+i√3)/2

The two complex roots are frequently referred to as ω and ω²

There is one real root and two imaginary roots.

x^3 = 1
x^3 - 1 = 0
(x - 1)(x^2 + x + 1) = 0

Set each factor to 0,
x - 1 = 0 ---> x = 1 (Obvious answer)

x^2 + x + 1 = 0

Using quadratic formula:
x = [-1 ± √(1 - 4)]/2
x = [-1 ± √(-3)]/2
x = (-1 ± i√3)/2

where i = √(-1)

The three solutions are:
x = 1
x = (-1 + i√3)/2
x = (-1 - i√3)/2

x^3 = 1
x = ³√1
x = 1

Solve X 3 1

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Do you mean 3^(x^3) = (1/27)^9? 3^(x^3) = (3^(-3))^9 3^(x^3) = 3^(-27) x^3 = -27 x = -3

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RE:
how to solve X^3=1 X=?
i heard that you can get 2 answer for X one is real number and another is imaginary number

x^3=1
x^3 - 1 = 0
(x - 1)(x^2 + x + 1) = 0

x = 1 or x = (-1 + isqrt(3))/2 or x = (-1 - isqrt(3))/2

x^3=1
x^3 - 1 = 0
(x - 1)(x^2 + x + 1) = 0

x = 1 or x = (1 + isqrt(3))/2 or x = (1 - isqrt(3))/2

To find the third roots of one, its x= exp(i * 2pi / 3 [since third roots] * k ). k = 0,1,2 OR 1,2,3 if you perfer. For k=0 or k=3, you get same number.
I should note that it could be x= exp(i * (2pi / 3 * k +2pi*n)). Where n is an integer. This 2pi*n of course comes from the fact that the exp function is periodic

yeah but for the REAL set of numbers, x = 1