Question Number 146377 by mathdanisur last updated on 13/Jul/21
$${Determine}\:{all}\:{pairs}\:{of}\:{real}\:{numbers} \\ $$$$\left({a};{b}\right)\:{such}\:{that}: \\ $$$${a}^{\mathrm{6}} −{b}^{\mathrm{4}} ={b}^{\mathrm{6}} −{a}^{\mathrm{4}} =\mathrm{4} \\ $$
Commented by mr W last updated on 13/Jul/21
$${a}={b}=\pm\sqrt{\mathrm{2}} \\ $$
Commented by mathdanisur last updated on 13/Jul/21
$${thankyou}\:{Ser},\:{solution} \\ $$
Commented by mr W last updated on 13/Jul/21
$${x}={a}^{\mathrm{2}} \\ $$$${y}={b}^{\mathrm{2}} \\ $$$${x}^{\mathrm{3}} −{y}^{\mathrm{2}} ={y}^{\mathrm{3}} −{x}^{\mathrm{2}} =\mathrm{4} \\ $$$${x}^{\mathrm{3}} +{x}^{\mathrm{2}} ={y}^{\mathrm{3}} +{y}^{\mathrm{2}} \\ $$$$\Rightarrow{x}={y} \\ $$$${x}^{\mathrm{3}} −{x}^{\mathrm{2}} =\mathrm{4} \\ $$$${x}^{\mathrm{2}} \left({x}−\mathrm{1}\right)=\mathrm{4} \\ $$$$\Rightarrow{x}=\mathrm{2}\Rightarrow{a}=\pm\sqrt{\mathrm{2}} \\ $$$$\Rightarrow{y}=\mathrm{2}\Rightarrow{b}=\pm\sqrt{\mathrm{2}} \\ $$
Commented by mathdanisur last updated on 13/Jul/21
$${cool}\:{Ser}\:{thanks},\:{y}=\pm\sqrt{\mathrm{2}}\Rightarrow{b}=\pm\sqrt{\mathrm{2}}.? \\ $$