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Question Number 148890 by mathdanisur last updated on 01/Aug/21

$$ifx+\frac{1}{x}=\sqrt{3}$$$$find{x}^{18}+x^{12}+x^6+1=?$$

Answered by nimnim last updated on 01/Aug/21

$$\mathrm{x}+\frac{1}{\mathrm{x}}=\sqrt{3}\Rightarrow {\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}=1$$$$\Rightarrow {\mathrm{x}}^6+\frac{1}{{\mathrm{x}}^6}+{3\mathrm{x}}^2.\frac{1}{{\mathrm{x}}^2}({\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2})=1\Rightarrow {\mathrm{x}}^6+\frac{1}{{\mathrm{x}}^6}=-2$$$$\Rightarrow {\mathrm{x}}^{12}+{2\mathrm{x}}^6+1=0\Rightarrow {({\mathrm{x}}^6+1)}^2=0\Rightarrow {\mathrm{x}}^6=-1$$$$\mathrm{Now},{\mathrm{x}}^{18}+{\mathrm{x}}^{12}+{\mathrm{x}}^6+1={(-1)}^3+{(-1)}^2-1+1$$$$=-1+1-1+1$$$$=0★$$

Commented by mathdanisur last updated on 01/Aug/21

$$ThankyouSer$$

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