\(x^3+y^3+z^3=3xyz\\ \Leftrightarrow\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\\ \Rightarrow x=y=z\)
tháy vào M ta có:
\(M=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\\ M=\left(1+1\right)\left(1+1\right)\left(1+1\right)=\left(1+1\right)^3=8\)
\(x^3+y^3+z^3=3xyz\Leftrightarrow x+y+z=0\)(chỗ này nếu k hiểu thì hỏi mk giải thích cho)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=-z\\y+z=-x\\z+x=-y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}z=-x-y\\x=-y-z\\y=-z-x\end{matrix}\right.\)
Thay vào M ta có:
\(M=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\)
\(=\left(1+\dfrac{-y-z}{y}\right)\left(1+\dfrac{-z-x}{z}\right)\left(1+\dfrac{-x-y}{x}\right)\)
\(=\left(1-1-\dfrac{z}{y}\right)\left(1-1-\dfrac{x}{z}\right)\left(1-1-\dfrac{y}{x}\right)\)
\(=\left(-\dfrac{z}{y}\right)\left(-\dfrac{x}{z}\right)\left(-\dfrac{y}{x}\right)\)
\(=-1\)