Yes the mistake is in the exponent##(e^{3i\pi}+e^{\frac{\pi}{3}})##. It's ##(e^{3i\pi}+e^{\frac{i\pi}{3}})##.
I just plugged ##\sqrt[3\,]{e^{\frac {2i\pi}{3}}} +\sqrt[3\,]{-\sqrt 3 e^{i\pi/6}}= 3^{1/6} e^{-5i\pi/18}+e^{\frac {4i\pi}{18}}## into ##\sqrt[3\,]...
##(-\frac 32 - i \frac{\sqrt 3}{2})^{1/3}= \sqrt[3\,] {e^{3i\pi}-e^{\frac{\pi}{3}}}## and ##(-1)^{2\over 9}=\sqrt[3\,] {e^{3i\pi}+e^{\frac{\pi}{3}}}##
When trying to solve ##(-3/2 - \frac{i}{2} \sqrt{3})=(a+bi)^3## I get ##a(a^2-b-2b^2)=\frac{3}{2}## and ##b(2a^2+a-b^2)=\frac{\sqrt{3}}{2}##.
How...
I don't know how to start with the factorization.
$$\frac{(-1)^{2/9} + (-3/2 - \frac{i}{2} \sqrt{3})^{(1/3)}}{(-1)^{2/9}- (-3/2 - \frac{i}{2} \sqrt{3})^{(1/3)}}$$
Any hints would be nice. Thank you.
"The group can, however, be given a different topology, in which the distance between two points is defined as the length of the shortest path in the group joining to "
What does this give us in formal math ?