Let
#x^3= 64(cos(pi)+isin(pi))#
#=>x^3= 64(-1+ixx0)#
#=>x^3+64=0#
#=>x^3+4^3=0#
#=>(x+4)(x^2-4x+16)=0#
So when #x+4=0#
#=>x=-4=4(-1+ixx0)=4(cos(pi)+isin(pi))#
when
# (x^2-4x+16)=0#
#=>x=(4pmsqrt(4^2-4*1*16))/2#
#=>x=(4pmsqrt(-48))/2#
#=>x=(4pm4sqrt(-3))/2#
#=>x=(2pmi2sqrt(3))#
#=>x=4(1/2pmisqrt(3)/2)#
So #=>x=4(cos(pi/3)pmisin(pi/3))#
So three cube roots are
#4(cos(pi)+isin(pi)),4(cos(pi/3)pmisin(pi/3))#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Applying de moivre's formula
we have
#x^3= 64(cos(pi)+isin(pi))#
#x= 64^(1/3)(cos((pi+2pik)/3)+isin((pi+2pik)/3))#
where k varies over the integer values from 0 to 2
For #k=0#
#x= 64^(1/3)(cos((pi+2pixx0)/3)+isin((pi+2pixx0)/3))#
#=>x= 4(cos((pi)/3)+isin((pi)/3))#
For #k=1#
#=>x= 64^(1/3)(cos((pi+2pixx1)/3)+isin((pi+2pixx1)/3))#
#=>x= 4(cos((pi)+isin((pi))#
For #k=2#
#x= 64^(1/3)(cos((pi+2pixx2)/3)+isin((pi+2pixx2)/3))#
#=>x= 4(cos(2pi-pi/3)+isin(2pi-pi/3))#
#=>x= 4(cos(pi/3)-isin(pi/3))#
Please note
A modest extension of the version of de Moivre's formula
https://en.wikipedia.org/wiki/De_Moivre%27s_formula
