If $x={\log }_{a}bc,y={\log }_{b}ca,z={\log }_{c}ab$, then $xyz=$

$x={\log }_{a}bc\Rightarrow bc=a^x\Rightarrow a^{x+1}=abc\Rightarrow a={\left(abc\right)}^{\frac{1}{x+1}}$

$y={\log }_{b}ac\Rightarrow ac=b^y\Rightarrow b^{y+1}=abc\Rightarrow b={\left(abc\right)}^{\frac{1}{y+1}}$

$z={\log }_{c}ab\Rightarrow ab=c^z\Rightarrow c^{z+1}=abc\Rightarrow c={\left(abc\right)}^{\frac{1}{z+1}}$

Now $abc={\left(abc\right)}^{\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}}$

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

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

$xyz=x+y+z+2$

Hence $xyz=x+y+z+2$