\(\left\{{}\begin{matrix}\left(x+1\right)\left(x^2+1\right)=y^3+1\\\left(y+1\right)\left(y^2+1\right)=z^3+1\\\left(z+1\right)\left(z^2+1\right)=x^3+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^3+x^2+x=y^3\left(1\right)\\y^3+y^2+y=z^3\\z^3+z^2+z=x^3\end{matrix}\right.\)

Giả sử \(x>y\Rightarrow x^3+x^2+x>y^3+y^2+y\)

\(\Rightarrow y^3>z^3\Leftrightarrow y>z\left(2\right)\)

\(\Rightarrow y^3+y^2+y>z^3+z^2+z\Rightarrow z>x\left(3\right)\)

Từ \(\left(2\right);\left(3\right)\Rightarrow y>x\) (Vô lí)

Giả sử \(x< y\Rightarrow x^3+x^2+x< y^3+y^2+y\)

\(\Rightarrow y^3< z^3\Leftrightarrow y< z\left(4\right)\)

\(\Rightarrow y^3+y^2+y< z^3+z^2+z\Rightarrow z< x\left(5\right)\)

Từ \(\left(4\right);\left(5\right)\Rightarrow y< x\) (Vô lí)

\(\Rightarrow x=y=z\)

\(\left(1\right)\Leftrightarrow x^3+x^2+x=x^3\)

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow x=y=z=0\) hoặc \(x=y=z=-1\)

a: Sửa đề: 

\(\left\{{}\begin{matrix}3xy=2\left(x+y\right)\\4yz=3\left(y+z\right)\\5xz=6\left(z+x\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+y}{xy}=\dfrac{3}{2}\\\dfrac{y+z}{yz}=\dfrac{4}{3}\\\dfrac{x+z}{xz}=\dfrac{5}{6}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{3}{2}\\\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{4}{3}\\\dfrac{1}{x}+\dfrac{1}{z}=\dfrac{5}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{3}{2}\\\dfrac{1}{y}=1\\\dfrac{1}{z}=\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow x=\dfrac{2}{3};y=1;z=3\)

b: Áp dụng tính chất của dãy tỉ số bằng nhau,ta được:

\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{9}=\dfrac{7x-3y+2z}{7\cdot4-3\cdot3+2\cdot9}=\dfrac{37}{37}=1\)

=>x=4; y=3; z=9