Question Number 159189 by mathlove last updated on 14/Nov/21
Answered by Rasheed.Sindhi last updated on 16/Nov/21
$${Let}\:\frac{{A}}{{B}}={x} \\ $$$${x}+\frac{\mathrm{1}}{{x}}=\mathrm{1};\:{x}^{\mathrm{2019}} +{x}^{−\mathrm{2019}} =? \\ $$$$\blacktriangleright{x}+\frac{\mathrm{1}}{{x}}=\mathrm{1} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −{x}+\mathrm{1}=\mathrm{0}\Rightarrow\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{3}} +\mathrm{1}=\mathrm{0}\Rightarrow{x}^{\mathrm{3}} =−\mathrm{1} \\ $$$$\therefore\:{x}\:{is}\:{cuberoot}\:{of}\:−\mathrm{1} \\ $$$$\therefore\:{x}=−\mathrm{1},−\omega,−\omega^{\mathrm{2}} \\ $$$$−\mathrm{1}\:{is}\:{a}\:{root}\:{of}\:{x}+\mathrm{1}=\mathrm{0} \\ $$$$−\omega,−\omega^{\mathrm{2}} \:{are}\:{the}\:{roots}\:{of}\:{x}^{\mathrm{2}} −{x}+\mathrm{1}=\mathrm{0} \\ $$$${where}\:\omega\:\:{is}\:{cuberoot}\:{of}\:{unity}. \\ $$$$\blacktriangleright{x}^{\mathrm{2019}} +{x}^{−\mathrm{2019}} =\left(−\omega\right)^{\mathrm{2019}} +\left(−\omega\right)^{−\mathrm{2019}} \\ $$$$−\left(\omega^{\mathrm{2019}} +\omega^{−\mathrm{2019}} \right)=−\left(\left\{\left(\omega\right)^{\mathrm{3}} \right\}^{\mathrm{673}} +\left\{\left(\omega\right)^{\mathrm{3}} \right\}^{−\mathrm{673}} \right) \\ $$$$=−\left(\mathrm{1}^{\mathrm{673}} +\mathrm{1}^{\mathrm{673}} \right)=−\mathrm{2} \\ $$$$\left(\frac{{A}}{{B}}\right)^{\mathrm{2019}} +\left(\frac{{B}}{{A}}\right)^{\mathrm{2019}} =−\mathrm{2} \\ $$