利用1的立方根,求下列实数的立方根 （1）1/8 （2）-27 计算（-1/2-二分之根号三*i)的八次方

(1)x^3-1=0,
设x1、x2、x3是1的三个立方根,
(x-1)(x^2+x+1)=0,
x1=1,
x2=-1/2+√3i/2,
x3=-1/2-√3i/2,
（2）(1/8)^(1/3),
x1=1/2,
x2=(1/2)(-1/2+√3i/2)=-1/4+√3i/4,
x3=(1/2)(-1/2-√3i/2))=-1/4-√3i/4,
(-27)^(1/3),
x1=-3,
x2=(-3)(-1/2+√3i/2)=3/2-3√3i/2,
x3=(-3)(-1/2-√3i/2)=3/2+3√3i/2,
（3）（-1/2-√3i/2）^8=（-1/2-√3i/2）^3*（-1/2-√3i/2）^3*（-1/2-√3i/2）^2
=1*1*(-1/2+√3i/2)=-1/2+√3i/2,
若用棣美弗定理,化成三角式,
（-1/2-√3i/2）^8=(cos4π/3+isin4π/3)^8
=cos32π/3+isin32π/3
=cos(10π+2π/3)+isin(10π+2π/3)
=cos2π/3+isin2π/3
=-1/2+√3i/2.