How do you solve x³ = 1?

      x³ = 1

  x³ - 1 = 0

Factor the difference of cubes:

(x - 1)(x² + x + 1) = 0

Set each factor equal to 0

Setting first factor = 0

x - 1 = 0

    x = 1  (that's ONE solution).

Setting second factor = 0

         x² + x + 1 = 0 

Use the quadratic formula:
                  ______ 
            -b ± Öb²-4ac
        x = —————————————
                2a 

where a = 1; b = 1; c = 1

                     _____________
             -(1) ± Ö(1)²-4(1)(1)
        x = ————————————————————————
                     2(1) 
                   _____ 
             -1 ± Ö1-4
        x = —————————————
                  2

                   __ 
             -1 ± Ö-3
        x = ———————————
                 2  

                    _ 
             -1 ± iÖ3
        x = ———————————
                 2 

To write these in the form A ± Bi
                     _
             -1     Ö3
        x = ———— ± ————·i 
              2      2

So there are three solutions:

(1)     x = 1

                     _
             -1     Ö3
(2)     x = ———— + ————·i 
              2      2
 
                     _
             -1     Ö3
(3)     x = ———— - ————·i 
              2      2

Edwin