\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
Với \(a+b+c=0\)
Làm nốt lười quá
Có :\(a^3+b^3+c^3=3abc\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
Xét \(a+b+c=0\)\(\Rightarrow\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{cases}}\) \(\Rightarrow\hept{\begin{cases}-a=b+c\\-b=c+a\\-c=a+b\end{cases}}\)
\(\Rightarrow A=\left(\frac{b+a}{b}\right)\left(\frac{c+b}{c}\right)\left(\frac{a+c}{a}\right)=\frac{-c}{b}.\frac{-a}{c}.\frac{-b}{a}=-1\)
Xét \(\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Rightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
mà \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0-\forall a,b,c\in R\)
\(\Rightarrow\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Rightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\Rightarrow a=b=c\)
\(\Rightarrow A=\left(1+\frac{a}{a}\right)\left(1+\frac{b}{b}\right)\left(1+\frac{c}{c}\right)\)
\(=\left(1+1\right)\left(1+1\right)\left(1+1\right)=2.2.2=8\)
Vậy\(A=-1\)hoặc\(A=8\)
