Question

If $a^{x-1}=bc,b^{y-1}=ac,c^{z-1}=ab$, then prove that$xy+yz+zx-xyz=0$.

Answer 1

We have,

$a^{x-1}=bc$

$a^x=abc$

$a={\left(abc\right)}^{1/x}$

Similarly,

$b={\left(abc\right)}^{1/y}$

$c={\left(abc\right)}^{1/z}$

Therefore,

$abc={\left(abc\right)}^{1/x}.{\left(abc\right)}^{1/y}.{\left(abc\right)}^{1/z}$

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

$\therefore 1/x+1/y+1/z=1$

$\frac{xy+yz+zx}{xyz}=1$

$xy+yz+zx=xyz$

$xy+yz+zx-xyz=0$

Hence, this is the answer.