Question

已知a^3+b^3+c^3=3abc,证明a=b=c

Answer 1

题目有错吧~

条件是不是少了~

少了a,b,c都是正数~

证:

a^3+b^3+c^3-3abc=0

(a+b)(a^2-ab+b^2)+c(c^2-3ab) =0

(a+b)(a^2-ab+b^2)+c(c^2-3ab+a^2-ab+b^2-a^2+ab-b^2)=0

(a+b)(a^2-ab+b^2)+c[(c^2-a^2-2ab-b^2)+(a^2-ab+b^2=0

(a+b)(a^2-ab+b^2)+c[c^2-(a+b)^2]+c(a^2-ab+b^2)=0(a+b+c)(a^2-ab+b^2)+c(a+b+c)(c-a-b)=0

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

(a+b+c)[1/2(a-b)^2+1/2(b-c)^2+1/2(a-c)^2]=0

∴a+b+c=0或a=b=c

∵a,b,c为正数

∴a=b=c