已知a,b,c为正数,且 a^2+b^3+c^3=3abc求证a=b=c

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[1/2(a-b)^2+1/2(b-c)^2+1/2(a-c)^2]=0
∵a,b,c为正数
∴a=b=c

a^3+b^3+c^3-3abc =0
=(a^3+3a^2b+3ab^2+b^3+c^3)-(3abc+3a^2b+3ab^2)
=[(a+b)^3+c^3]-3ab(a+b+c)
=(a+b+c)(a^2+b^2+2ab-ac-bc+c^2)-3ab(a+b+c)
=(a+b+c)(a^2+b^2+c^2+2ab-3ab-ac-bc)
=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)
=1/2 *(a+b+c)[(a-b)^2+(a-c)^2+(b-c)^2]
=0
a,b,c为正数，所以a+b+c >0
所以(a-b)^2+(a-c)^2+(b-c)^2=0
所以 a=b=c