关于三次函数的题

hbghlyj

设$f(x)=x^3+ax^2+bx+c,g(x)=x^3+p x^2+qx+r(a,b,c,p,q,r\in\mathbb C)$.复数$\alpha,\beta,\gamma$为f(x)的零点.满足$\{g(\alpha),g(\beta),g(\gamma)\}=\{A,B,C\}$的$(a_i,b_i,c_i)$共有27组.求证:$\sum a_i=27 p$,$\sum b_i= 9 p^2$,$\sum c_i= p^3$.

hbghlyj

先用MMA验证一下:

Clear[p,q,r];
{g\[Alpha], g\[Beta],g\[Gamma]} = #^3 + p #^2 + q # + r & /@ {\[Alpha], \[Beta], \[Gamma]};
sy = SymmetricPolynomial[#, {g\[Alpha], g\[Beta], g\[Gamma]}] & /@ {1,2, 3};
T = Simplify[First[SymmetricReduction[#, {\[Alpha], \[Beta], \[Gamma]}, {-a,b, -c}]] & /@ sy];
p = RandomReal[]; q = RandomReal[]; r = RandomReal[];
Chop[Total[{a, b, c} /. Solve[T == {RandomReal[], RandomReal[], RandomReal[]}, {a, b,c}]] - {27 p, 9 p^2, p^3}] == {0, 0, 0}

输出
True