Question

If $a^{x-1}=bc,b^{y-1}=ac,c^{z-1}=ab$, then find the value of $xy+yz+zx-xyz$.

Answer 1

We have,

$a^{x-1}=bc$

$\frac{a^x}{a}=abc$

$a^x=abc$

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

Similarly,

$b={\left(abc\right)}^{1/y}$ $\dots \left(2\right)$

$c={\left(abc\right)}^{1/z}$ $\dots \left(3\right)$

Multiply $\left(1\right),\left(2\right),\left(3\right)$

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

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

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

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

$xy+yz+zx=xyz$

$xy+yz+zx-xyz=0$

Hence, this is the answer.