參考站長大的解法
https://math.pro/db/thread-407-1-1.html

小弟也來個解法。這個題目的數據設計得不好，要解三次方程

\(\begin{align}
  & x+xy+xyz=9 \\
& y+yz+xyz=14 \\
& z+zx+xyz=12 \\
&  \\
& xyz=t \\
& x+xy=9-t \\
& z\left( 1+x+xy \right)=12 \\
& z\left( 10-t \right)=12 \\
& z=\frac{12}{10-t},x=\frac{9}{15-t},y=\frac{14}{13-t} \\
&  \\
& \frac{9}{15-t}+\frac{9\times 14}{\left( 15-t \right)\left( 13-t \right)}+\frac{9\times 14\times 12}{\left( 15-t \right)\left( 13-t \right)\left( 10-t \right)}=9 \\
& \frac{1}{15-t}+\frac{14}{\left( 15-t \right)\left( 13-t \right)}+\frac{168}{\left( 15-t \right)\left( 13-t \right)\left( 10-t \right)}=1 \\
& {{t}^{3}}-37{{t}^{2}}+438t-1512=0 \\
& \left( t-6 \right)\left( {{t}^{2}}-31t+252 \right)=0 \\
& t=6 \\
&  \\
& x=1,y=2,z=3 \\
\end{align}\)