\(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)=0\\ \text{⇔}a^2b-a^2c+b^2a-b^2a+c^2a-c^2b=0\\ \text{⇔}\left(a^2b-ab^2-abc+b^2c\right)-\left(a^2c-abc-ac^2+bc^2\right)=0\\ \text{⇔}b\left(a^2-ab-ac+bc\right)-c\left(a^2-ab-ac+bc\right)=0\\ \text{⇔}\left(b-c\right)\left[a\left(a-b\right)-c\left(a-b\right)\right]=0\\ \text{⇔}\left(b-c\right)\left(a-b\right)\left(a-c\right)=0\\ \text{⇔}\left[{}\begin{matrix}b=c\\a=b\\a=c\end{matrix}\right.\)
Vậy, ba số a,b,c tồn tại hai số bằng nhau