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Question Number 66058 by arvinddayama01@gmail.comm last updated on 08/Aug/19

$$Ifx+\frac{1}{x}=1$$$$$$$$findoutvalue:-$$$$$$$$\frac{x^{20}+x^{17}+x^{14}+x^{11}}{x^{17}+x^{14}+x^{11}+x^8}=?$$

Commented by Prithwish sen last updated on 08/Aug/19

$${\mathrm{x}}^2-\mathrm{x}+1=0\Rightarrow (\mathrm{x}+1)({\mathrm{x}}^2-\mathrm{x}+1)=0\Rightarrow {\mathrm{x}}^3=-1$$$$\frac{{\mathrm{x}}^3({\mathrm{x}}^{17}+{\mathrm{x}}^{14}+{\mathrm{x}}^{11}+{\mathrm{x}}^8)}{{\mathrm{x}}^{17}+{\mathrm{x}}^{14}+{\mathrm{x}}^{11}+{\mathrm{x}}^8}={\mathrm{x}}^3=-1$$

Commented by $@ty@m123 last updated on 08/Aug/19

$$Ihaveadoubt:$$$$Ifyouwrite$$$${\mathrm{x}}^2-\mathrm{x}+1=0\Rightarrow (\mathrm{x}+1)({\mathrm{x}}^2-\mathrm{x}+1)=0\Rightarrow {\mathrm{x}}^3=-1$$$$thenyoucanalsowrite:$$$${\mathrm{x}}^2-\mathrm{x}+1=0\Rightarrow (\mathrm{x}+n)({\mathrm{x}}^2-\mathrm{x}+1)=0\Rightarrow x=-n\Rightarrow n\in R$$

Commented by Prithwish sen last updated on 08/Aug/19

$$\mathrm{I}\mathrm{have}\mathrm{just}\mathrm{used}\mathrm{the}\mathrm{formula}\mathrm{of}$$$${\mathrm{a}}^3+{\mathrm{b}}^3=(\mathrm{a}+\mathrm{b})({\mathrm{a}}^2-\mathrm{ab}+{\mathrm{b}}^2)\mathrm{where}\mathrm{I}\mathrm{just}$$$$\mathrm{put}\mathrm{a}=\mathrm{x}\mathrm{and}\mathrm{b}=1.$$

Commented by Prithwish sen last updated on 09/Aug/19

$$\mathrm{I}\mathrm{think}\mathrm{the}\mathrm{equation}{\mathrm{x}}^3+1=0\mathrm{has}3$$$$\mathrm{roots}\mathrm{they}\mathrm{are}-1,\frac{1\pm \sqrt{3}\mathrm{i}}{2.}$$$$\mathrm{Now}\mathrm{x}+1=0\mathrm{has}\mathrm{the}\mathrm{root}\mathrm{x}=-1$$$$\mathrm{and}{\mathrm{x}}^2-\mathrm{x}+1=0\mathrm{has}\mathrm{two}\mathrm{roots}\frac{1\pm \sqrt{3}\mathrm{i}}{2}$$$$\mathrm{just}\mathrm{like}\mathrm{the}\mathrm{equation}$$$${\mathrm{x}}^2-5\mathrm{x}+6=0\mathrm{has}2\mathrm{roots}2\mathrm{and}3$$$$\mathrm{of}\mathrm{which}\mathrm{x}-2=0\mathrm{is}\mathrm{satisfied}\mathrm{by}\mathrm{x}=2$$$$\mathrm{and}\mathrm{x}-3=0\mathrm{is}\mathrm{satisfied}\mathrm{by}\mathrm{x}=3.\mathrm{But}\mathrm{as}\mathrm{a}\mathrm{total}$$$$\mathrm{the}\mathrm{entire}\mathrm{equation}\mathrm{is}\mathrm{satisfied}\mathrm{by}\mathrm{both}2$$$$\mathrm{roots}.$$

Commented by $@ty@m123 last updated on 09/Aug/19

$$Onemoreobservation:$$$$Thegivenexpression\ne x^3$$$$whenx=-1$$

Answered by $@ty@m123 last updated on 08/Aug/19

$$x+\frac{1}{x}=1$$$$x^2-x+1=0$$$$x=\frac{1\pm \sqrt{3}i}{2}$$$$Thegivenexpression=x^3$$$$={\left(\frac{1\pm \sqrt{3}i}{2}\right)}^3$$$$=\frac{1}{8}\{1\pm {\left(\sqrt{3}i\right)}^3\pm 3\sqrt{3}i+3.(-3)\}$$$$=\frac{1}{8}(1\mp 3\sqrt{3}i\mp 3\sqrt{3}i-9)$$$$=\frac{-8}{8}$$$$=-1$$

Answered by afjyormathchamp@gmail.com last updated on 08/Aug/19

$$\therefore x^2-x+1=0$$$$(x+1)(x^2-x+1)=0x^3=-1$$$$\Rightarrow \frac{x^3(x^{17}+x^{14}+x^{11}+x^8)}{(x^{17}+x^{14}+x^{11}+x^8)}=x^3=-1$$

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