\(a^3+b^3+c^3=3abc\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Rightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Rightarrow\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}a+b+c=0\\a=b=c=0\end{cases}}\)
*) Xét a+b+c=0 suy ra \(\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\). Khi đó
\(M=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{c+a}{a}=\frac{-c}{b}\cdot\frac{-a}{c}\cdot\frac{-b}{a}=-1\)
*)Xét a=b=c=0 loại vì mẫu khác 0