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If $x + \frac{1}{x} = 1$ Then find the value of $p$, where $p = x^{4000} + \frac{1}{x^{4000}}.$

I tried to solve it by squaring the equation. But by this method , i can get the value of $x^{4096} + \frac{1}{x^{4096}}$ But not the value of $p$.

How I can solve this? Thanks in advance.

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    We know $x + \frac{1}{x} = 1$, so $x^2 + 1 = x,$ or $x^2 -x + 1 = 0.$ Solve for $x,$ and substitute in $p.$2012-03-17

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Hint:

  • $\dfrac{x^2+ 1}{x}=1 \implies x^2-x+1=0 \implies x^3+1=0~~~ \mbox{with $x \neq 1$}$

  • So, $x$ is a complex cube root of $-1$.

  • The user miracle173 points out that we don't have to recall Euler and De-Moivre as just plugging in $x^3=-1$ is sufficient. $x^{4000}+\dfrac{1}{x^{4000}}=-x-\dfrac 1 x=-1$

(Thanks for pointing that out, miracle173.)


  • Recall Euler's Form of a complex number and De-Moivre's Theorem
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    @miracle173 I have edited to add your point. How does it look? (Thanks $f$or telling anyway.)2012-03-17