The first thing to do is to put the number in the form of #rhoe^(thetai)#

#rho=sqrt((1/2)^2+(sqrt(3)/2)^2)=sqrt(1/4+3/4)=1#

#theta=arctan((sqrt(3)/2)/(-1/2))=arctan(-sqrt(3))=-pi/3+kpi#. Let's choose #(2pi)/3#since we are in the second quadrant. Pay attention that #-pi/3# is in the fourth quadrant, and this is wrong.

Your number is now:

#1e^((2pii)/3)#

Now the roots are:

#root(3)(1)e^(((2kpi+(2pi)/3)i)/3), k in ZZ#

#=e^((((6kpi+2pi)i)/9), k in ZZ#

so you can choose k= 0, 1, 2 and obtain:

#e^((2pii)/9#, #e^((8kpii)/9# and #e^((14kpii)/9#

or #cos((2pi)/9)+ isin((2pi)/9)#, #cos((8pi)/9)+ isin((8pi)/9)# and

#cos((14pi)/9)+ isin((14pi)/9)#.

For me this is a dead end, because I cannot compute trigonometric functions of multiples of #pi/9#. We must rely on a calculator:

#0.7660+0.6428i#

#-0.9397+0.3420i#

#0.1736-0.9848i#