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ab+bc+ac=0,則2(ab+bc+ac)=0 ab+bc+ab+ac+bc+ac=0 (a+c)b+(b+c)a+(a+b)c=0 a+b+c=0,則a+b=-c,a+c=-b,b+c=-a -b^2-a^2-c^2=0 即a^2+b^2+c^2=0 a=b=c=0
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