假設 A+B+C+D = 0 請證明 (ABC+BCD+CDA+DAB)^2 = (BC-AD)(CA-BD)(AB-CD)有點難，不過希望大家能幫忙小弟！

熱心網友

ABC+BCD+CDA+DAB = AB(C+D)+CD(A+B)=AB(C+D)-CD(C+D)=(AB-CD)(C+D)....(1)ABC+BCD+CDA+DAB = AC(B+D)+BD(A+C)=AC(B+D)-BD(B+D)=(AC-BD)(B+D)....(2)(1)*(2): (ABC+BCD+CDA+DAB)^2 = (AB-CD)(C+D)(AC-BD)(B+D)= (AB-CD)(AC-BD)(BC+BD+CD+D^2)= (AB-CD)(AC-BD)[BC+BD+D(C+D)]= (AB-CD)(AC-BD)(BC+BD-D(A+B)]= (AB-CD)(AC-BD)(BC-DA)證畢

熱心網友

設(X-A)(X-B)(X-C)(X-D)=X^4-a1X^3+a2X^2-a3X+a4由于(BC-AD)(CA-BD)(AB-CD)為A,B,C,D的對稱多項式，則其可寫成a1，a2，a3，a4的多項式，并計算(BC-AD)(CA-BD)(AB-CD)=-(a1)^2*a4+(a3)^2,當a1=0時，(BC-AD)(CA-BD)(AB-CD)=(a3)^2。

熱心網友

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