\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow c^3+\left(a+b\right)^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)^3-3c\left(a+b\right)\left(a+b+c\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab+2bc+2ca-3ca-3bc-3ab\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ca-bc-ab\right)=0\)
\(\Leftrightarrow\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
Mà a, b ,c đôi một khác nhau nên \(a+b+c=0\)
Khi đó \(\frac{1}{a^2+b^2-c^2}=\frac{1}{a^2+b^2-\left(-a-b\right)^2}=-\frac{1}{2ab}\)
Tương tự \(\frac{1}{b^2+c^2-a^2}=-\frac{1}{2bc};\frac{1}{c^2+a^2-b^2}=-\frac{1}{2ca}\)
\(\Rightarrow A=-\left(\frac{1}{2ab}+\frac{1}{2bc}+\frac{1}{2ca}\right)=-\frac{a+b+c}{2abc}=0\)
