Ta có: \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow a^3+3ab\left(a+b\right)+b^3+c^3-3abc-3ab\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ab-ac+c^2\right)-3ab\left(a+b+c\right)=0\) \(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-ac-bc=0\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
+) a = b = c
\(\Rightarrow P=2+2+2=6\)
+) a + b + c = 0
\(\Rightarrow\left\{\begin{matrix}a=-b-c\\b=-a-c\\c=-a-b\end{matrix}\right.\)
\(\Rightarrow P=\left(\frac{-b-c}{b}+1\right)\left(\frac{-a-c}{c}+1\right)\left(\frac{-a-b}{a}+1\right)\)
\(=\frac{-c}{b}\times\frac{-a}{c}\times\frac{-b}{a}\)
\(=-1\)
Từ \(a^3+b^3+c^3=3abc\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Rightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Rightarrow\left[\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{matrix}\right.\)\(\Rightarrow\left[\begin{matrix}a+b+c=0\\a=b=c=0\end{matrix}\right.\)
*)Xét \(a+b+c=0\) suy ra \(\left\{\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\). Khi đó:
\(P=\left(\frac{a}{b}+1\right)\left(\frac{b}{c}+1\right)\left(\frac{c}{a}+1\right)=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{c+a}{a}=\frac{-c}{b}\cdot\frac{-a}{c}\cdot\frac{-b}{a}=-1\)
*)Xét \(a=b=c=0\) loại vì mẫu khác 0
