设x^6=-1即x^6=cos[(2n+1)π]+isin[(2n+1)π]
所以x=cos[(2n+1)π/6]+isin[(2n+1)π/6];
n=0时,x1=cos(π/6)+isin(π/6)=√3/2+i/2;
n=1时,x2=cos(π/2)+isin(π/2)=i;
n=2时,x3=cos(5π/6)=isin(5π/6)=-√3/2+i/2;
n=3时,x4=cos(7π/6)+isin(7π/6)=-√3/2-i/2;
n=4时,x5=cos(3π/2)+isin(3π/2)=-i,
n=5时,x6=cos(11π/6)+isin(11π/6)=√3/2-i/2;