Question Number 212531 by RojaTaniya last updated on 16/Oct/24
Answered by a.lgnaoui last updated on 17/Oct/24
![x=1+^7 (√2) +(^7 (√2) )^2 +(^7 (√2) )^3 +(^7 (√(2 )) )^4 +(^7 (√2) )^5 +((√2) )^6 =((1−(^7 (√2) )^7 )/(1−(^7 (√2) )))=(1/(^7 (√2) −1))⇒ (1/x)=^7 (√2) −1 (1/x^3 )−(3/x)=(1/x)((1/x^2 )−3)= (^7 (√2) −1)[(^7 (√2) )−1)^2 −3] (1/x^3 )−(3/x)=(^7 (√2) )^3 −3(^7 (√2) )^2 +2](https://www.tinkutara.com/question/Q212547.png)
$$\boldsymbol{\mathrm{x}}=\mathrm{1}+^{\mathrm{7}} \sqrt{\mathrm{2}}\:+\left(^{\mathrm{7}} \sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} +\left(^{\mathrm{7}} \sqrt{\mathrm{2}}\:\right)^{\mathrm{3}} +\left(^{\mathrm{7}} \sqrt{\mathrm{2}\:}\:\right)^{\mathrm{4}} +\left(^{\mathrm{7}} \sqrt{\mathrm{2}}\:\right)^{\mathrm{5}} +\left(\sqrt{\mathrm{2}}\:\right)^{\mathrm{6}} \\ $$$$=\frac{\mathrm{1}−\left(^{\mathrm{7}} \sqrt{\mathrm{2}}\:\right)^{\mathrm{7}} }{\mathrm{1}−\left(^{\mathrm{7}} \sqrt{\mathrm{2}}\:\right)}=\frac{\mathrm{1}}{\:^{\mathrm{7}} \sqrt{\mathrm{2}}\:−\mathrm{1}}\Rightarrow\:\:\:\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}=^{\mathrm{7}} \sqrt{\mathrm{2}}\:−\mathrm{1} \\ $$$$\:\:\: \\ $$$$\left.\:\:\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}^{\mathrm{3}} }−\frac{\mathrm{3}}{\boldsymbol{\mathrm{x}}}=\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\left(\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }−\mathrm{3}\right)=\:\:\left(^{\mathrm{7}} \sqrt{\mathrm{2}}\:−\mathrm{1}\right)\left[\left(^{\mathrm{7}} \sqrt{\mathrm{2}}\:\right)−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{3}\right] \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}^{\mathrm{3}} }−\frac{\mathrm{3}}{\boldsymbol{\mathrm{x}}}=\left(^{\mathrm{7}} \sqrt{\mathrm{2}}\:\right)^{\mathrm{3}} −\mathrm{3}\left(^{\mathrm{7}} \sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} +\mathrm{2} \\ $$$$ \\ $$