Ta có: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\)
\(=2a^2+2b^2+2c^2-2ab-2bc-2ac\)
\(=2\left(a^2+b^2+c^2+2ab+2ac+2bc\right)-6ab-6bc-6ac\)
\(=2\left(a+b+c\right)^2-6\left(ab+bc+ac\right)\)
\(=2.6^2-6.12=0\)
Mà : \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(a-c\right)^2\ge0\)
nên \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
Do đó: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
<=> \(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(a-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}}\Leftrightarrow a=b=c\)
Vậy \(\left(a-b\right)^{2012}+\left(b-c\right)^{2013}+\left(c-a\right)^{2014}=0\)