If $(xy)^{a-1}=z$

$(yz)^{b-1}=x$

$(zx)^{c-1}=y$

then $ab+bc+ca=?$

$(xy)^{a-1}=z$

$(yz)^{b-1}=x$

$(zx)^{c-1}=y$

Multiplying the equations we get,

$(xy)^{a-1}\times(yz)^{b-1}\times(zx)^{c-1}$=$x y z$

Comparing we get,

$a+c-2=1\Rightarrow a+c=3$

$a+b-2=1\Rightarrow a+b=3$

$b+c-2=1\Rightarrow b+c=3$

By solving these eqautions we get, $a=b=c=\dfrac{3}{2}$

Therefore,$ab+bc+ca=\dfrac { 3 }{ 2 } \times \dfrac { 3 }{ 2 } +\dfrac { 3 }{ 2 } \times \dfrac { 3 }{ 2 } +\dfrac { 3 }{ 2 } \times \dfrac { 3 }{ 2 } =\dfrac { 27 }{ 4 }$