Question Number 30455 by ajfour last updated on 22/Feb/18
Commented by mrW2 last updated on 25/Feb/18
$${p}=\frac{{ac}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)}{{a}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)+{c}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)} \\ $$$${q}=\frac{{c}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)}{{a}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)+{c}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)} \\ $$$$\frac{{p}}{{q}}=\frac{{a}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)}{{c}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)} \\ $$$${please}\:{confirm}. \\ $$
Commented by ajfour last updated on 25/Feb/18
$${yes}\:{sir},\:{with}\:{some}\:{effort}\:{i}\:{obtained} \\ $$$${the}\:{same}. \\ $$
Commented by ajfour last updated on 25/Feb/18
Commented by ajfour last updated on 25/Feb/18