Alg3.3.15

If $a^{x-1}=bc,b^{y-1}=ca,c^{z-1}=ab\,$ show that $xy+yz+zx=xyz\,$

Let $a^{x-1}=bc,b^{y-1}=ca,c^{z-1}=ab\,$

Simplifying

$\frac{a^x}{a}=bc,\frac{b^y}{b}=ca,\frac{c^z}{c}=ab\,$

$a^x=abc,a^y=abc,a^z=abc\,$

Hence $a^x=b^y=c^z=abc\,$

Therefore

$a=(abc)^{\frac{1}{x}},b=(abc)^{\frac{1}{y}},c=(abc)^{\frac{1}{z}}\,$

Multiplying the three,we get

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

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

$xyz=xy+yz+zx\,$