[Jom ~]

เอาคูณกันหมดก็จะได้

$xyz+x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{xyz} = ABC$

$xyz+(x+\frac{1}{y})+(y+\frac{1}{z})+(z+\frac{1}{x})+\frac{1}{xyz} = ABC$

$xyz+A+B+C+\frac{1}{xyz} = ABC$

$xyz+\frac{1}{xyz} = ABC-(A+B+C)$

^^