Question

Answer 1

\(\left(a+b+c\right)^2+12=4\left(a+b+c\right)+2\left(ab+bc+ac\right)\)

=>\(a^2+b^2+c^2+2\left(ab+bc+ac\right)+12=4\left(a+b+c\right)+2\left(ab+bc+ac\right)\)

=>\(a^2+b^2+c^2+12-4\left(a+b+c\right)=0\)

=>\(\left(a^2-4a+4\right)+\left(b^2-4b+4\right)+\left(c^2-4c+4\right)=0\)

=>\(\left(a-2\right)^2+\left(b-2\right)^2+\left(c-2\right)^2=0\)

=>\(\left\{{}\begin{matrix}a-2=0\\b-2=0\\c-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=2\\c=2\end{matrix}\right.\)

=>a=b=c=2