Question Number 12386 by Joel576 last updated on 21/Apr/17
$${f}\left({x}\:+\:\frac{\mathrm{1}}{{x}}\right)\:=\:\frac{{x}^{\mathrm{6}} \:+\:\mathrm{1}}{\mathrm{27}} \\ $$$${f}\left({x}\right)\:=\:? \\ $$
Answered by ajfour last updated on 21/Apr/17
![let (x+(1/x))=t since (x+(1/x))^3 =x^3 +(1/x^3 )+3(x+(1/x)) t^3 = ((x^6 +1)/x^3 ) +3t ⇒ (t^3 −3t) = (((x^6 +1))/( (√((x^6 +1)−1)))) ⇒(t^3 −3t)^2 {(x^6 +1)−1}= (x^6 +1)^2 (x^6 +1)^2 −(t^3 −3t)^2 (x^6 +1)+(t^3 −3t)^2 =0 (x^6 +1)= (((t^3 −3t)^2 ±(√((t^3 −3t)^4 −4(t^3 −3t)^2 )))/2) f(t)=(((x^6 +1))/(27))=(((t^3 −3t)^2 ±(√((t^3 −3t)^4 −4(t^3 −3t)^2 )))/(54)) f(x)=(((x^3 −3x)^2 )/(54))[1±(√(1−(4/((x^3 −3x)^2 )))) ] .](https://www.tinkutara.com/question/Q12393.png)
$${let}\:\:\left({x}+\frac{\mathrm{1}}{{x}}\right)={t} \\ $$$${since}\:\:\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} ={x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\mathrm{3}\left({x}+\frac{\mathrm{1}}{{x}}\right) \\ $$$$\:{t}^{\mathrm{3}} \:=\:\frac{{x}^{\mathrm{6}} +\mathrm{1}}{{x}^{\mathrm{3}} }\:+\mathrm{3}{t} \\ $$$$\Rightarrow\:\left({t}^{\mathrm{3}} −\mathrm{3}{t}\right)\:=\:\frac{\left({x}^{\mathrm{6}} +\mathrm{1}\right)}{\:\sqrt{\left({x}^{\mathrm{6}} +\mathrm{1}\right)−\mathrm{1}}} \\ $$$$\Rightarrow\left({t}^{\mathrm{3}} −\mathrm{3}{t}\right)^{\mathrm{2}} \left\{\left({x}^{\mathrm{6}} +\mathrm{1}\right)−\mathrm{1}\right\}=\:\left({x}^{\mathrm{6}} +\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\left({x}^{\mathrm{6}} +\mathrm{1}\right)^{\mathrm{2}} −\left({t}^{\mathrm{3}} −\mathrm{3}{t}\right)^{\mathrm{2}} \left({x}^{\mathrm{6}} +\mathrm{1}\right)+\left({t}^{\mathrm{3}} −\mathrm{3}{t}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({x}^{\mathrm{6}} +\mathrm{1}\right)=\:\frac{\left({t}^{\mathrm{3}} −\mathrm{3}{t}\right)^{\mathrm{2}} \pm\sqrt{\left({t}^{\mathrm{3}} −\mathrm{3}{t}\right)^{\mathrm{4}} −\mathrm{4}\left({t}^{\mathrm{3}} −\mathrm{3}{t}\right)^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\boldsymbol{{f}}\left(\boldsymbol{{t}}\right)=\frac{\left(\boldsymbol{{x}}^{\mathrm{6}} +\mathrm{1}\right)}{\mathrm{27}}=\frac{\left(\boldsymbol{{t}}^{\mathrm{3}} −\mathrm{3}\boldsymbol{{t}}\right)^{\mathrm{2}} \pm\sqrt{\left(\boldsymbol{{t}}^{\mathrm{3}} −\mathrm{3}\boldsymbol{{t}}\right)^{\mathrm{4}} −\mathrm{4}\left(\boldsymbol{{t}}^{\mathrm{3}} −\mathrm{3}\boldsymbol{{t}}\right)^{\mathrm{2}} }}{\mathrm{54}} \\ $$$$\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\frac{\left(\boldsymbol{{x}}^{\mathrm{3}} −\mathrm{3}\boldsymbol{{x}}\right)^{\mathrm{2}} }{\mathrm{54}}\left[\mathrm{1}\pm\sqrt{\mathrm{1}−\frac{\mathrm{4}}{\left(\boldsymbol{{x}}^{\mathrm{3}} −\mathrm{3}\boldsymbol{{x}}\right)^{\mathrm{2}} }}\:\right]\:. \\ $$
Commented by ajfour last updated on 21/Apr/17
$${for}\:{x}=\mathrm{1} \\ $$$$\:\:{f}\left(\mathrm{1}+\mathrm{1}\right)=\frac{\mathrm{1}+\mathrm{1}}{\mathrm{27}}=\:\frac{\mathrm{2}}{\mathrm{27}}\:. \\ $$
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 21/Apr/17
= =(t^2 −2)t^2 x^2 −(t^2 −2)tx−(t^2 −1)tx+(t^2 −1)= =t^2 (t^2 −2)(tx−1)−[(t^2 −2+t^2 −1)]tx+(t^2 −1)= (t^4 −2t^2 −2t^2 +3)tx−(t^4 −2t^2 −t^2 +1)⇒ x^6 =(1/2)t(t^4 −4t^2 +3)(t±(√(t^2 −4)))−(t^4 −3t^2 +1) x^6 +1=(1/2)t(t^2 −3)[(t^2 −1)(t±(√(t^2 −4)))−2t] f(x)=((x(x^2 −3))/(54))[(x^2 −1)(x±(√(x^2 −4)))−2x]](https://www.tinkutara.com/question/Q12411.png)
$${x}+{x}^{−\mathrm{1}} ={t}\Rightarrow{x}^{\mathrm{2}} −{tx}+\mathrm{1}=\mathrm{0}\Rightarrow{x}=\frac{{t}\pm\sqrt{{t}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} ={tx}−\mathrm{1} \\ $$$${x}^{\mathrm{3}} ={tx}^{\mathrm{2}} −{x}={t}\left({tx}−\mathrm{1}\right)−{x}=\left({t}^{\mathrm{2}} −\mathrm{1}\right){x}−{t} \\ $$$${x}^{\mathrm{4}} =\left({t}^{\mathrm{2}} −\mathrm{1}\right){x}^{\mathrm{2}} −{tx}=\left({t}^{\mathrm{2}} −\mathrm{1}\right)\left({tx}−\mathrm{1}\right)−{tx}= \\ $$$$=\left({t}^{\mathrm{2}} −\mathrm{2}\right){tx}−\left({t}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$${x}^{\mathrm{6}} ={x}^{\mathrm{4}} ×{x}^{\mathrm{2}} =\left[\left({t}^{\mathrm{2}} −\mathrm{2}\right){tx}−\left({t}^{\mathrm{2}} −\mathrm{1}\right)\right]\left({tx}−\mathrm{1}\right)= \\ $$$$=\left({t}^{\mathrm{2}} −\mathrm{2}\right){t}^{\mathrm{2}} {x}^{\mathrm{2}} −\left({t}^{\mathrm{2}} −\mathrm{2}\right){tx}−\left({t}^{\mathrm{2}} −\mathrm{1}\right){tx}+\left({t}^{\mathrm{2}} −\mathrm{1}\right)= \\ $$$$={t}^{\mathrm{2}} \left({t}^{\mathrm{2}} −\mathrm{2}\right)\left({tx}−\mathrm{1}\right)−\left[\left({t}^{\mathrm{2}} −\mathrm{2}+{t}^{\mathrm{2}} −\mathrm{1}\right)\right]{tx}+\left({t}^{\mathrm{2}} −\mathrm{1}\right)= \\ $$$$\left({t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{2}} +\mathrm{3}\right){tx}−\left({t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{2}} −{t}^{\mathrm{2}} +\mathrm{1}\right)\Rightarrow \\ $$$${x}^{\mathrm{6}} =\frac{\mathrm{1}}{\mathrm{2}}{t}\left({t}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{2}} +\mathrm{3}\right)\left({t}\pm\sqrt{{t}^{\mathrm{2}} −\mathrm{4}}\right)−\left({t}^{\mathrm{4}} −\mathrm{3}{t}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$${x}^{\mathrm{6}} +\mathrm{1}=\frac{\mathrm{1}}{\mathrm{2}}{t}\left({t}^{\mathrm{2}} −\mathrm{3}\right)\left[\left({t}^{\mathrm{2}} −\mathrm{1}\right)\left({t}\pm\sqrt{{t}^{\mathrm{2}} −\mathrm{4}}\right)−\mathrm{2}{t}\right] \\ $$$${f}\left({x}\right)=\frac{{x}\left({x}^{\mathrm{2}} −\mathrm{3}\right)}{\mathrm{54}}\left[\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left({x}\pm\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}\right)−\mathrm{2}{x}\right] \\ $$