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Question Number 184260 by Ml last updated on 04/Jan/23

$$\mathrm{x}+\frac{1}{\mathrm{x}}=-1$$$${\mathrm{x}}^{1377}=?$$

Answered by SEKRET last updated on 04/Jan/23

$${\mathbf{x}}^2+1=-\mathbf{x}$$$${\mathbf{x}}^2+\mathbf{x}+1=0$$$$(\mathbf{x}-1)({\mathbf{x}}^2+\mathbf{x}+1)=0$$$${\mathbf{x}}^3-1=0$$$${\mathbf{x}}^3=1$$$${\mathbf{x}}^{1377}={\left({\mathbf{x}}^3\right)}^{459}=1$$$$$$

Answered by Rasheed.Sindhi last updated on 04/Jan/23

$$x^2+x+1=0$$$$x^2=-x-1$$$$x^3=-x^2-x=-(-x-1)-x=1$$$$x^{1377}={\left(x^3\right)}^{459}={\left(1\right)}^{459}=1$$

Answered by Rasheed.Sindhi last updated on 04/Jan/23

$$x^2+x+1=0\Rightarrow x=\omega ,{\omega }^2$$$$\bullet x=\omega :x^{1377}={\left(\omega \right)}^{1377}={\left({\omega }^3\right)}^{459}$$$$={\left(1\right)}^{459}=1$$$$\bullet x={\omega }^2:x^{1377}={\left({\omega }^2\right)}^{3\times 459}$$$$={\left({\omega }^3\right)}^{2\times 459}={\left(1\right)}^{2\times 459}=1$$

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