Ta có: \(A=\left(\frac{a}{b}+1\right)\left(\frac{b}{c}+1\right)\left(\frac{c}{a}+1\right)\)
\(=\left(\frac{a}{b}+\frac{b}{b}\right)\cdot\left(\frac{b}{c}+\frac{c}{c}\right)\cdot\left(\frac{c}{a}+\frac{a}{a}\right)\)
\(=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)
Ta có: \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
Trường hợp 1: a+b+c=0
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)
\(\Leftrightarrow A=\frac{-c\cdot\left(-b\right)\cdot\left(-a\right)}{abc}=\frac{-abc}{abc}=-1\)
Trường hợp 2: \(a^2+b^2+c^2-ab-bc-ac=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+a^2-2ac+c^2=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Leftrightarrow a=b=c\)
\(\Leftrightarrow A=\frac{2a\cdot2b\cdot2c}{abc}=\frac{8abc}{abc}=8\)
