ĐK: \(abc\ne0\)
\(a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2+c^2-c\left(a+b\right)\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
TH1: \(a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\a+c=-b\end{matrix}\right.\)
\(A=\left(\frac{a+b}{b}\right)\left(\frac{b+c}{c}\right)\left(\frac{a+c}{a}\right)=\frac{\left(-a\right).\left(-b\right)\left(-c\right)}{abc}=-1\)
TH2: \(a=b=c\)
\(\Rightarrow A=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
