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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
