What is the cubed root of -1? Is it just -1 or actually +-1? Thx!?

Answer 1

It's -1.
There are also two imaginary (complex) roots.

Answer 2

Since -1 squared is + 1 and -1 cubed is -1, the cube root of - 1 is -1. It cannot be + or - 1.

Answer 3

There are actually three cube roots of - 1. They are -1, 1/2(1 + i√3), and 1/2(1 - i√3), where
i ≡ √-1. We usually use only - 1, and ignore the other two.

check:
(1/2)(1 + i√3)(1/2)(1 + i√3)(1/2)(1 + i√3) =
(1/8)(1 + i2√3 - 3)(1 + i√3) =
(1/8)(i2√3 - 2)(1 + i√3) =
(1/4)(i√3 - 1)(1 + i√3) =
- (1/4)(1 - i√3)(1 + i√3) =
- (1/4)(1(1 + i√3) - i√3(1 + i√3)) =
- (1/4)(1 + i√3 - (i√3 - 3)) =
- (1/4)(1 + i√3 - i√3 + 3) =
- (1/4)(4) = - 1

The other root checks out in the same manner.

Answer 4

-1

because if you are going to trace its cube:


-1(-1)= +1 (-1) = -1

Answer 5

-1 = exp(ip(1 + 2n))
(-1)^(1/3) = exp(ipi/3 + 2nipi/3)
=cos(pi/3 + 2npi/3) + isin(pi/3 + 2npi/3)

Three unique values for n=0, 1 and 2.

n=0
cos(pi/3) + i sin(pi/3) = (1 + sqrt(3)i)/2

n=1
cos(pi) + isin(pi) = -1

n=2
cos(5pi/3) + isin(5pi/3) = cos(pi/3) - isin(pi/3) = (1 - sqrt(3)i)/2

Those are your three cube roots of -1. Only -1 is real, the other two are complex.

Answer 6

(-1)^2= 1
(-1)^3= -1
__ __
V-1 = V-1
__ __
3V-1 = 3V(-1)^3
= (-1)^3*1/3
= -1
Solution:
__
3V-1 = -1

Answer 7

ITS STILL -1

-1X-1= -1X-1 ONCE AGAIN STILL EQUALS -1