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
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![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](/uploads/image/z/206880-24-0.jpg?t=%E5%88%A9%E7%94%A81%E7%9A%84%E7%AB%8B%E6%96%B9%E6%A0%B9%2C%E6%B1%82%E4%B8%8B%E5%88%97%E5%AE%9E%E6%95%B0%E7%9A%84%E7%AB%8B%E6%96%B9%E6%A0%B9+%EF%BC%881%EF%BC%891%2F8+%EF%BC%882%EF%BC%89-27+%E8%AE%A1%E7%AE%97%EF%BC%88-1%2F2-%E4%BA%8C%E5%88%86%E4%B9%8B%E6%A0%B9%E5%8F%B7%E4%B8%89%2Ai%29%E7%9A%84%E5%85%AB%E6%AC%A1%E6%96%B9)
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
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
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.
设Z是-27的立方根
z^3=-27 z^3/-27=1 (z/-3)^3=1
∵1的立方根有1,-1/2±根号3/2i
∴z/-3=-1/2±根号3/2i
z=3/2±3根号3/2i
1/8也是这么做