\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Leftrightarrow\)\(\dfrac{1}{2}\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-2ac-2bc\right)=0\)
\(\Leftrightarrow\dfrac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\right]=0\)
\(\Rightarrow\)\(\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
TH1: \(a+b+c=0\Rightarrow a=-\left(b+c\right);b=-\left(a+c\right);c=-\left(a+b\right)\)
\(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
\(M=\left(1+\dfrac{-b-c}{b}\right)\left(1+\dfrac{-c-a}{c}\right)\left(1+\dfrac{-a-b}{a}\right)\)
\(M=\left(1-1-\dfrac{c}{b}\right)\left(1-1-\dfrac{a}{c}\right)\left(1-1-\dfrac{b}{a}\right)\)
\(M=\left(-\dfrac{c}{b}\right)\left(-\dfrac{a}{c}\right)\left(-\dfrac{b}{a}\right)=-1\)
TH2: \(a=b=c\)
\(M=\left(1+\dfrac{a}{a}\right)\left(1+\dfrac{a}{a}\right)\left(1+\dfrac{a}{a}\right)=2.2.2=8\)
Do \(a^3+b^3+c^3=3abc\).
Nên ta dễ dàng cm đc: \(a+b+c=0\)
\(\Rightarrow\)a + b = -c; b+c = -a; a + c = -b (1)
\(\Rightarrow M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
=\(\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{a+c}{a}\)(2)
Thay (1) vào (2) được:
\(M=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}=-1\)
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