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

2个回答

  • (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.