2015/079) if $a^{x-1} = bc, b^{y-1} = ca, c^{z-1} = ab$ show that $xy + yz + zx = xyz$

$a^{x-1} = bc$

hence $a^x = abc$

or $a = (abc)^{\frac{1}{x}}\cdots (1)$

similarly

$b = (abc)^{\frac{1}{y}}\cdots (2)$

$c = (abc)^{\frac{1}{z}}\cdots (3)$ 

hence $abc = (abc)^{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$

hence 

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

or $yz + zx + xy = xyz$

proved