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Question Number 8073 by 314159 last updated on 29/Sep/16

Commented by 123456 last updated on 30/Sep/16

$$\cosh \left(x\right)=\frac{e^x+e^{-x}}{2}$$$$x^4+x^{-4}=47$$$$x=e^t$$$$e^{4t}+e^{-4t}=47$$$$2\cdot \frac{e^{4t}+e^{-4t}}{2}=47$$$$2\cosh \left(4t\right)=47$$$$t=\frac{\mathrm{argcosh}\left(\frac{47}{2}\right)}{4}$$$$x^3+x^{-3}=2\cdot \frac{e^{3t}+e^{-3t}}{2}=2\cosh \left(3t\right)$$$$=2\cosh \left(\frac{3\mathrm{argcosh}\left(\frac{47}{2}\right)}{4}\right)$$

Commented by Rasheed Soomro last updated on 30/Sep/16

$$Appreciationsforthisnovelapproach!$$$$Canweobtaintheanswerinmoresimplifiedform?$$

Commented by prakash jain last updated on 30/Sep/16

$$\mathrm{This}\mathrm{gives}\mathrm{only}\mathrm{one}\mathrm{solution}.$$$$x^4+\frac{1}{x^4}\mathrm{gives}8\mathrm{solutions}\mathrm{which}\mathrm{are}\mathrm{be}$$$$\alpha ,\beta ,\gamma ,\theta ,\frac{1}{\alpha },\frac{1}{\beta },\frac{1}{\gamma },\frac{1}{\theta }$$$$\mathrm{effectively}\mathrm{we}\mathrm{get}4\mathrm{solutions}\mathrm{for}{x}^3+\frac{1}{x^3}$$$$\mathrm{as}\mathrm{given}\mathrm{in}{\mathrm{Rasheed}}^{\prime }\mathrm{s}\mathrm{Answer}?\mathrm{How}\mathrm{do}$$$$\mathrm{we}\mathrm{get}\mathrm{same}4\mathrm{answer}\mathrm{using}\mathrm{above}\mathrm{approach}?$$$$\mathrm{Actual}\mathrm{calculation}\mathrm{of}\mathrm{the}\mathrm{value}\mathrm{gives}18.$$

Commented by 123456 last updated on 30/Sep/16

$$y=\cosh \left(t\right)=\frac{e^t+e^{-t}}{2}$$$$x=e^t$$$$2y=x+\frac{1}{x}$$$$2xy=x^2+1$$$$x^2-2xy+1=0$$$$\mathrm{\Delta }=4y^2-4=4(y^2-1)$$$$x=y\pm \sqrt{y^2-1}(x⩾0)$$$$e^t=y+\sqrt{y^2-1}$$$$t=\ln (y+\sqrt{y^2-1})$$$$\mathrm{argcosh}\left(t\right)=\ln (t+\sqrt{t^2-1})$$

Commented by 123456 last updated on 30/Sep/16

$$\mathrm{we}\mathrm{lost}\mathrm{some}\mathrm{roots}$$$$y=x^2\Rightarrow x=\pm \sqrt{y}$$$$\mathrm{also}$$$$\cos (x+2\pi k)=\cos \left(x\right)$$$$\cosh \left(xi\right)=\cos \left(x\right)$$$$\mathrm{so}\cosh \mathrm{are}\mathrm{periodic}\mathrm{too},\mathrm{but}\mathrm{the}\mathrm{period}$$$$\mathrm{is}2\pi i$$

Commented by prakash jain last updated on 30/Sep/16

$$\mathrm{Thanks}.\mathrm{I}\mathrm{understand}\mathrm{now}.$$$$2\cosh \left(4t\right)=47$$$$\mathrm{Will}give4t=2in\pi \pm {\cosh }^{-1}\left(47/2\right)$$$$\mathrm{So}\mathrm{we}\mathrm{get}8\mathrm{solution}\mathrm{for}t\mathrm{for}n=0,1,2,3$$$$$$

Answered by Rasheed Soomro last updated on 29/Sep/16

$$x^4+\frac{1}{x^4}=47,x^3+\frac{1}{x^3}=?$$$$x^4+\frac{1}{x^4}=47$$$$x^4+2+\frac{1}{x^4}=47+2$$$${\left(x^2\right)}^2+2\left(x^2\right)\left(\frac{1}{x^2}\right)+\left(\frac{1}{x^2}\right)=49$$$${(x^2+\frac{1}{x^2})}^2={(\pm 7)}^2$$$$x^2+\frac{1}{x^2}=7\mid x^2+\frac{1}{x^2}=-7$$$$x^2+2+\frac{1}{x^2}=7+2\mid x^2+2+\frac{1}{x^2}=-7+2$$$${\left(x\right)}^2+2\left(x\right)\left(\frac{1}{x}\right)+{\left(\frac{1}{x}\right)}^2=9\mid {\left(x\right)}^2+2\left(x\right)\left(\frac{1}{x}\right)+{\left(\frac{1}{x}\right)}^2=-5$$$${(x+\frac{1}{x})}^2={(\pm 3)}^2\mid {(x+\frac{1}{x})}^2={(\pm \sqrt{5}i)}^2$$$$x+\frac{1}{x}=\pm 3\mid x+\frac{1}{x}=\pm \sqrt{5}i$$$$$$$${(x+\frac{1}{x})}^3={(\pm 3)}^3\mid {(x+\frac{1}{x})}^3={(\pm \sqrt{5}i)}^3$$$$x^3+\frac{1}{x^3}+3(x+\frac{1}{x})={(\pm 3)}^2(\pm 3)\mid x^3+\frac{1}{x^3}+3(x+\frac{1}{x})={(\pm \sqrt{5}i)}^2(\pm \sqrt{5}i)$$$$x^3+\frac{1}{x^3}+3(\pm 3)=\left(9\right)(\pm 3)\mid x^3+\frac{1}{x^3}+3(\pm \sqrt{5}i)=(-5)(\pm \sqrt{5}i)$$$$x^3+\frac{1}{x^3}=\left(9\right)(\pm 3)-3(\pm 3)\mid x^3+\frac{1}{x^3}=(-5)(\pm \sqrt{5}i)-3(\pm \sqrt{5}i)$$$$x^3+\frac{1}{x^3}=(9-3)(\pm 3)\mid x^3+\frac{1}{x^3}=(-5-3)(\pm \sqrt{5}i)$$$$x^3+\frac{1}{x^3}=6(\pm 3)\mid x^3+\frac{1}{x^3}=-8(\pm \sqrt{5}i)$$$$x^3+\frac{1}{x^3}=\pm 18\mid x^3+\frac{1}{x^3}=\mp 8\sqrt{5}i$$$$x^3+\frac{1}{x^3}=18,-18,-8\sqrt{5}i,8\sqrt{5}i$$

Answered by prof.kerdus last updated on 29/Sep/16

$$its18$$$$\bullet ⌣\bullet$$

Answered by Rasheed Soomro last updated on 30/Sep/16

$$\underset{-}{\mathbf{Another}\mathbf{approach}}$$$$x^4+\frac{1}{x^4}=47,x^3+\frac{1}{x^3}=?$$$${(x^4+\frac{1}{x^4})}^3={\left(47\right)}^3$$$${(a+\frac{1}{a})}^3=a^3+\frac{1}{a^3}+3(a+\frac{1}{a})$$$${\left(x^4\right)}^3+{\left(\frac{1}{x^4}\right)}^3+3(x^4+\frac{1}{x^4})=103823$$$$x^{12}+\frac{1}{x^{12}}+3\left(47\right)=103823$$$$x^{12}+\frac{1}{x^{12}}=103823-141=103682$$$$$$$${\left(x^6\right)}^2+{\left(\frac{1}{x^6}\right)}^2=103682$$$${\left(x^6\right)}^2+2+{\left(\frac{1}{x^6}\right)}^2=103682+2$$$${\left(x^6\right)}^2+2\left(x^6\right)\left(\frac{1}{x^6}\right)+{\left(\frac{1}{x^6}\right)}^2=103684$$$${(x^6+\frac{1}{x^6})}^2={(\pm 322)}^2$$$$x^6+\frac{1}{x^6}=\pm 322$$$$x^6+2+\frac{1}{x^6}=\pm 322+2$$$${\left(x^3\right)}^2+2\left(x^3\right)\left(\frac{1}{x^3}\right)+{\left(\frac{1}{x^3}\right)}^2=324,-320$$$${(x^3+\frac{1}{x^3})}^2={(\pm 18)}^2\mid {(x^3+\frac{1}{x^3})}^2={(\pm 8\sqrt{5}i)}^2$$$$x^3+\frac{1}{x^3}=\pm 18\mid x^3+\frac{1}{x^3}=\pm 8\sqrt{5}i$$$$x^3+\frac{1}{x^3}=18,-18,8\sqrt{5}i,-8\sqrt{5}i$$

Answered by Rasheed Soomro last updated on 30/Sep/16

$$Slightlydefferentversion$$$$ofmyanswerdated29/9/16$$$$x^4+\frac{1}{x^4}=47,x^3+\frac{1}{x^3}=?$$$$Letx+\frac{1}{x}=y$$$${(x+\frac{1}{x})}^2=y^2$$$$x^2+\frac{1}{x^2}=y^2-2$$$${(x^2+\frac{1}{x^2})}^2={(y^2-2)}^2$$$$x^4+\frac{1}{x^4}={(y^2-2)}^2-2=47$$$${(y^2-2)}^2=47+2=49$$$${(y^2-2)}^2={(\pm 7)}^2$$$$y^2-2=\pm 7$$$$y^2=\pm 7+2$$$$y^2=7+2,-7+2$$$$y^2={(\pm 3)}^2\mid y^2={(\pm \sqrt{-5})}^2$$$$y=3,-3\mid y=\sqrt{5}i,-\sqrt{5}i$$$$x+\frac{1}{x}=3,-3,\sqrt{5}i,-\sqrt{5}i$$$$Forremainingprocessseemyanswerdated29/09/16$$

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