2019/006) if $a^{x-1} = bc, b^{y-1} = ca, c^{z-1} = ab$ show that $xy + yz + zx = xyz$

From $a^{(x-1)} = bc$ we have $a^x= abc$ or $a=(abc)^\frac{1}{x}\cdots(1)$

similarly $b=(abc)^\frac{1}{y}\cdots(2)$
$c=(abc)^\frac{1}{z}\cdots(3)$

multiplying (1) (2) and (3) we get $abc = (abc)^{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}$
or

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

multiplying both sides by xyz we get the result