let 'x' represent the a number whose sixth power is equal to 64

x^6 = 64

x^6 - 64 = 0

factoring the equation we can write:

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

click here to see the step by step factoring:

the linear factors give two solutions:

x1 = 2

x2 = -2

solving the quadratic factors we find the complex roots:

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

x3 = -1 - i√3

x4 = -1 + i√3

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

x5 = 1 - i√3

x6 = 1 + i√3

the complex numbers whose sixth power is equal to 64 are:

1 + i√3, -1 + i√3, 1 - i√3, -1 - i√3