Question Number 90406 by I want to learn more last updated on 23/Apr/20
$$\mathrm{If}\:\:\:\:\:\:\mathrm{x}\:\:−\:\:\frac{\mathrm{1}}{\mathrm{x}}\:\:\:=\:\:\:\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{4}} \:\:−\:\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }\:\:\:=\:\:\:??? \\ $$
Answered by som(math1967) last updated on 23/Apr/20
$${x}−\frac{\mathrm{1}}{{x}}=\mathrm{3} \\ $$$$\Rightarrow\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} =\mathrm{9} \\ $$$$\Rightarrow\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{4}{x}.\frac{\mathrm{1}}{{x}}=\mathrm{9} \\ $$$$\Rightarrow\left({x}+\frac{\mathrm{1}}{{x}}\right)=\pm\sqrt{\mathrm{13}}\: \\ $$$${again}\:\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} =\mathrm{9} \\ $$$$\therefore{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{9}+\mathrm{2}=\mathrm{11} \\ $$$${x}^{\mathrm{4}} −\frac{\mathrm{1}}{{x}^{\mathrm{4}} }=\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\left({x}+\frac{\mathrm{1}}{{x}}\right)\left({x}−\frac{\mathrm{1}}{{x}}\right) \\ $$$$=\pm\mathrm{33}\sqrt{\mathrm{13}}\:{ans} \\ $$
Commented by I want to learn more last updated on 23/Apr/20
$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir} \\ $$
Answered by MJS last updated on 23/Apr/20
$${x}={u}+{v}=\frac{\left({u}+{v}\right)^{\mathrm{2}} \left({u}−{v}\right)}{\left({u}+{v}\right)\left({u}−{v}\right)} \\ $$$$\frac{\mathrm{1}}{{x}}=\frac{{u}−{v}}{\left({u}+{v}\right)\left({u}−{v}\right)} \\ $$$${x}^{\mathrm{4}} −\frac{\mathrm{1}}{{x}^{\mathrm{4}} }=\frac{\left({u}+{v}\right)^{\mathrm{8}} \left({u}−{v}\right)^{\mathrm{4}} −\left({u}−{v}\right)^{\mathrm{4}} }{\left({u}+{v}\right)^{\mathrm{4}} \left({u}−{v}\right)^{\mathrm{4}} }= \\ $$$$=\frac{\left({u}^{\mathrm{2}} −{v}^{\mathrm{2}} \right)^{\mathrm{4}} \left({u}+{v}\right)^{\mathrm{4}} −\left({u}−{v}\right)^{\mathrm{4}} }{\left({u}^{\mathrm{2}} −{v}^{\mathrm{2}} \right)^{\mathrm{4}} } \\ $$$${x}−\frac{\mathrm{1}}{{x}}=\mathrm{3}\:\Rightarrow\:{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{1}=\mathrm{0}\:\Rightarrow\:{x}=\frac{\mathrm{3}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$$${u}=\frac{\mathrm{3}}{\mathrm{2}}\wedge{v}^{\mathrm{2}} =\frac{\mathrm{13}}{\mathrm{4}} \\ $$$$\frac{\left(\frac{\mathrm{9}}{\mathrm{4}}−\frac{\mathrm{13}}{\mathrm{4}}\right)^{\mathrm{4}} \left({u}+{v}\right)^{\mathrm{4}} −\left({u}−{v}\right)^{\mathrm{4}} }{\left(\frac{\mathrm{9}}{\mathrm{4}}−\frac{\mathrm{13}}{\mathrm{4}}\right)^{\mathrm{4}} }= \\ $$$$=\left({u}+{v}\right)^{\mathrm{4}} −\left({u}−{v}\right)^{\mathrm{4}} =\mathrm{8}{uv}\left({u}^{\mathrm{2}} +{v}^{\mathrm{2}} \right)= \\ $$$$=\mathrm{8}{uv}\left(\frac{\mathrm{9}}{\mathrm{4}}+\frac{\mathrm{13}}{\mathrm{4}}\right)=\mathrm{44}{uv}=\pm\mathrm{33}\sqrt{\mathrm{13}} \\ $$
Commented by I want to learn more last updated on 23/Apr/20
$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir} \\ $$