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
