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Question Number 213636 by hardmath last updated on 11/Nov/24

$${\mathrm{x}}^2-\mathrm{x}+1=0$$$$\mathrm{find}:{\mathrm{x}}^{10}=?$$

Answered by A5T last updated on 11/Nov/24

$$x^2=x-1\Rightarrow x^3=x^2-x=-1\Rightarrow x^9=-1\Rightarrow x^{10}=-x$$$$x=\frac{1\underset{-}{+}\sqrt{1-4}}{2}\Rightarrow -x=\frac{-1}{2}\stackrel{-}{+}\frac{i\sqrt{3}}{2}$$

Answered by Frix last updated on 12/Nov/24

$$x^2-x+1=0$$$$\mathrm{Borrow}x=-1$$$$(x+1)(x^2-x+1)=0$$$$x^3+1=0$$$$x=-1\left(\mathrm{Return}\mathrm{with}\mathrm{thanks}\right)$$$$x={\mathrm{e}}^{\pm \mathrm{i}\frac{\pi }{3}}$$$$\Rightarrow x^{10}={\mathrm{e}}^{\pm \frac{10\pi }{3}}={\mathrm{e}}^{\mp \mathrm{i}\frac{2\pi }{3}}=-x$$

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