\(a^2+b^2+c^2-ab-bc-ca=0\\ \)
Nhân cả 2 vế với 2 ta được
\(2a^2+2b^2+2c^2-2ab-2bc-2ac=0\\=> \left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)=0\\=> \left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\\ =>\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(a-c\right)^2=0\\\left(b-c\right)^2=0\end{matrix}\right.=>\left\{{}\begin{matrix}a=b\\a=c\\b=c\end{matrix}\right.=>a=b=c\)