infinity)-(-1)^(1/3)=(cos 180-isin180)^(1/3)=cos (180+360k)/3-isin(180+360k)/3k=0,1,......
r ( cos Θ + I sin Θ)
it is an alternative way for expressing a complex number called
"Polar Form for a Complex Number"
In fact, the polar form of a complex number and a concept
called DeMoivre's Theorem provide the usual tools we use
for figuring out the values complex bases raised to powers
named with real valued exponents.
powers.
There is only one answer. For example a^(1/2) means the principal square root of a.
For this question Google gives the correct answer. Also, (-1)^(1/3) isn't -1. It is a complex number. Wolfram alpha gives the correct value of it and it is the same value as i^(2/3). So that is where the errors happened, once you start considering complex numbers, (-1)^(1/3) is no longer -1.
i = e^(iπ/2)
so i^(2/3) = (e^(iπ/2))^(2/3) = e^(iπ/3) = cos(π/3) + i*sin(π/3)
Note that I used radians since I'm used to using them while the other answers used degrees.
I got the answer -1 so my belief their is only one solution.
It pretty much saying "(-1)^(1/3)^3 = -1, and (-1)^3 = -1."
https://www.google.com/#fp=92d015eca11a9…
The fact is (-1)^[1/(odd)] is always going to equal (-1) as long
the odd number is a whole number so (-infinity, infinity)
