Does the equality (X at the 3rd power = 1) have any solutions except 1 & -1 on the set of the complex numbers?

Question

Shame on me for that -1 :D. I'm too tired, but of corse I know this is not a good explanation.
Thanks for everyone : ).

Answer 1

#1: -1 is not a solution of x³=1, since (-1)³=-1≠1
#2: This has three solutions: 1, (-1+i√3)/2, and (-1-i√3)/2. You may verify this by simple multiplication.

Answer 2

you are asking for cube roots of 1
if you know trigonometry use this equation
x^1/n = x^1/n(cos(2*k*pi/n)+i sin(2*k*pi/n))
x^1/n is the root you are looking for on the left side
on the right side it is the main root, in your example it was 1.
pi=3.14159.....
n=root you are finding, in you example it was 3
k=any number less than k
so for x= 1 and n=3, substitue 0 ,1 and 2 in the equation for k,
when it is 0: cos 2*0*pi/n=1 +i sin 2*0*k/n=0 and so the answer is
1(1+0) or just 1
im too lazy to do the math for k=1 and k=2

Answer 3

Only solution is "1". "-1" isn't valid since x^3 will equal -1 when x=-1.

Answer 4

No. If you substituted i or -i for x you wouldn't get 1 for a solution.
-1 isn't a solution either because (-1)^3 = -1.

Answer 5

x^3 = 1 has only solution, x=1, with multiplicity of 3

(if you don't know what multiplicity means, don't worry about it. bottom line is the only answer, even including the complex numbers, is x=1 for this equation)

Answer 6

x^3 = 1 has 3 solutions... in polar notation (radius,theta) they are (1,0) (1, 2pi/3) (1, 4pi/3).

-1 is not a solution.

Answer 7

x^3 - 1 = 0
(x-1)(x^2 + x + 1) = 0
x = (1, (-1+sqrt(-3))/2, (-1-sqrt(-3))/2)

Answer 8

NO.