Question Number 100976 by M±th+et+s last updated on 29/Jun/20
$${find}\:{all}\:{possible}\:{values}\:{of}\:{x},{y},{z}\:{in}\:{terms} \\ $$$${of}\:{a},{b},{c}\:{gor}\:{a}\:{triplet}\:\left({x},{y},{z}\right)\:{that}\:{satisfy} \\ $$$$ \\ $$$${x}+\frac{\mathrm{1}}{{y}}={a} \\ $$$$ \\ $$$${y}+\frac{\mathrm{1}}{{z}}={b} \\ $$$$ \\ $$$${z}+\frac{\mathrm{1}}{{x}}={c} \\ $$
Commented by john santu last updated on 29/Jun/20
$$\Rightarrow{x}\:=\:\frac{{ay}−\mathrm{1}}{{y}}\:\wedge\:\mathrm{y}\:=\:\frac{{bz}−\mathrm{1}}{{z}} \\ $$$$\Rightarrow\:{z}+\:\frac{{y}}{{ay}−\mathrm{1}}\:=\:{c}\:\wedge\:\mathrm{y}\:=\:\frac{{bz}−\mathrm{1}}{{z}} \\ $$$$\Rightarrow\:{z}\:+\:\frac{\frac{{bz}−\mathrm{1}}{{z}}}{{a}\left(\frac{{bz}−\mathrm{1}}{{z}}\right)−\mathrm{1}}\:=\:{c}\: \\ $$$${z}\:+\:\frac{{bz}−\mathrm{1}}{{abz}−{a}−{z}}\:=\:{c}\: \\ $$$${z}\:+\:\frac{{bz}−\mathrm{1}}{{z}\left({ab}−\mathrm{1}\right)−{a}}\:=\:{c}\: \\ $$$${z}\left({z}\left({ab}−\mathrm{1}\right)−{a}\right)+{bz}−\mathrm{1}\:=\:{c}\left({z}\left({ab}−\mathrm{1}\right)−{a}\right) \\ $$$$\left({ab}−\mathrm{1}\right){z}^{\mathrm{2}} +\left({b}−{a}\right){z}−\mathrm{1}=\left({abc}−{c}\right){z}−{ac} \\ $$$$\left({ab}−\mathrm{1}\right){z}^{\mathrm{2}} +\left({b}−{a}+{c}−{abc}\right){z}+{ac}−\mathrm{1}=\mathrm{0} \\ $$$${z}\:=\:\frac{{abc}+{a}−{b}−{c}\:\pm\:\sqrt{\left({b}−{a}+{c}−{abc}\right)^{\mathrm{2}} −\mathrm{4}\left({ac}−\mathrm{1}\right)\left({ab}−\mathrm{1}\right)}}{\mathrm{2}\left({ab}−\mathrm{1}\right)} \\ $$