Question

If $(xy{)}^{a-1}=z$
$(yz{)}^{b-1}=x$
$(zx{)}^{c-1}=y$
then $ab+bc+ca=?$

Answer 1

$(xy{)}^{a-1}=z$
$(yz{)}^{b-1}=x$
$(zx{)}^{c-1}=y$
Multiplying the equations we get,
$(xy{)}^{a-1}\times (yz{)}^{b-1}\times (zx{)}^{c-1}$=$xyz$
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, $a=b=c=\frac{3}{2}$
Therefore,$ab+bc+ca=\frac{3}{2}\times \frac{3}{2}+\frac{3}{2}\times \frac{3}{2}+\frac{3}{2}\times \frac{3}{2}=\frac{27}{4}$