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Question Number 90406 by I want to learn more last updated on 23/Apr/20

If x − (1/x) = 3 x^4 − (1/x^4 ) = ???

$$\mathrm{If}\:\:\:\:\:\:\mathrm{x}\:\:−\:\:\frac{\mathrm{1}}{\mathrm{x}}\:\:\:=\:\:\:\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{4}} \:\:−\:\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }\:\:\:=\:\:\:??? \\ $$

Answered by som(math1967) last updated on 23/Apr/20

x−(1/x)=3 ⇒(x−(1/x))^2 =9 ⇒(x+(1/x))^2 −4x.(1/x)=9 ⇒(x+(1/x))=±(√(13)) again (x−(1/x))^2 =9 ∴x^2 +(1/x^2 )=9+2=11 x^4 −(1/x^4 )=(x^2 +(1/x^2 ))(x+(1/x))(x−(1/x)) =±33(√(13)) ans

$${x}−\frac{\mathrm{1}}{{x}}=\mathrm{3} \\ $$$$\Rightarrow\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} =\mathrm{9} \\ $$$$\Rightarrow\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{4}{x}.\frac{\mathrm{1}}{{x}}=\mathrm{9} \\ $$$$\Rightarrow\left({x}+\frac{\mathrm{1}}{{x}}\right)=\pm\sqrt{\mathrm{13}}\: \\ $$$${again}\:\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} =\mathrm{9} \\ $$$$\therefore{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{9}+\mathrm{2}=\mathrm{11} \\ $$$${x}^{\mathrm{4}} −\frac{\mathrm{1}}{{x}^{\mathrm{4}} }=\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\left({x}+\frac{\mathrm{1}}{{x}}\right)\left({x}−\frac{\mathrm{1}}{{x}}\right) \\ $$$$=\pm\mathrm{33}\sqrt{\mathrm{13}}\:{ans} \\ $$

Commented by I want to learn more last updated on 23/Apr/20

I appreciate sir

$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir} \\ $$

Answered by MJS last updated on 23/Apr/20

x=u+v=(((u+v)^2 (u−v))/((u+v)(u−v))) (1/x)=((u−v)/((u+v)(u−v))) x^4 −(1/x^4 )=(((u+v)^8 (u−v)^4 −(u−v)^4 )/((u+v)^4 (u−v)^4 ))= =(((u^2 −v^2 )^4 (u+v)^4 −(u−v)^4 )/((u^2 −v^2 )^4 )) x−(1/x)=3 ⇒ x^2 −3x−1=0 ⇒ x=(3/2)±((√(13))/2) u=(3/2)∧v^2 =((13)/4) ((((9/4)−((13)/4))^4 (u+v)^4 −(u−v)^4 )/(((9/4)−((13)/4))^4 ))= =(u+v)^4 −(u−v)^4 =8uv(u^2 +v^2 )= =8uv((9/4)+((13)/4))=44uv=±33(√(13))

$${x}={u}+{v}=\frac{\left({u}+{v}\right)^{\mathrm{2}} \left({u}−{v}\right)}{\left({u}+{v}\right)\left({u}−{v}\right)} \\ $$$$\frac{\mathrm{1}}{{x}}=\frac{{u}−{v}}{\left({u}+{v}\right)\left({u}−{v}\right)} \\ $$$${x}^{\mathrm{4}} −\frac{\mathrm{1}}{{x}^{\mathrm{4}} }=\frac{\left({u}+{v}\right)^{\mathrm{8}} \left({u}−{v}\right)^{\mathrm{4}} −\left({u}−{v}\right)^{\mathrm{4}} }{\left({u}+{v}\right)^{\mathrm{4}} \left({u}−{v}\right)^{\mathrm{4}} }= \\ $$$$=\frac{\left({u}^{\mathrm{2}} −{v}^{\mathrm{2}} \right)^{\mathrm{4}} \left({u}+{v}\right)^{\mathrm{4}} −\left({u}−{v}\right)^{\mathrm{4}} }{\left({u}^{\mathrm{2}} −{v}^{\mathrm{2}} \right)^{\mathrm{4}} } \\ $$$${x}−\frac{\mathrm{1}}{{x}}=\mathrm{3}\:\Rightarrow\:{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{1}=\mathrm{0}\:\Rightarrow\:{x}=\frac{\mathrm{3}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$$${u}=\frac{\mathrm{3}}{\mathrm{2}}\wedge{v}^{\mathrm{2}} =\frac{\mathrm{13}}{\mathrm{4}} \\ $$$$\frac{\left(\frac{\mathrm{9}}{\mathrm{4}}−\frac{\mathrm{13}}{\mathrm{4}}\right)^{\mathrm{4}} \left({u}+{v}\right)^{\mathrm{4}} −\left({u}−{v}\right)^{\mathrm{4}} }{\left(\frac{\mathrm{9}}{\mathrm{4}}−\frac{\mathrm{13}}{\mathrm{4}}\right)^{\mathrm{4}} }= \\ $$$$=\left({u}+{v}\right)^{\mathrm{4}} −\left({u}−{v}\right)^{\mathrm{4}} =\mathrm{8}{uv}\left({u}^{\mathrm{2}} +{v}^{\mathrm{2}} \right)= \\ $$$$=\mathrm{8}{uv}\left(\frac{\mathrm{9}}{\mathrm{4}}+\frac{\mathrm{13}}{\mathrm{4}}\right)=\mathrm{44}{uv}=\pm\mathrm{33}\sqrt{\mathrm{13}} \\ $$

Commented by I want to learn more last updated on 23/Apr/20

I appreciate sir

$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir} \\ $$

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