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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 $p.$ – user2468 Mar 17 '12 at 5:06
up vote 5 down vote accepted


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

  • 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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If these hints are not sufficient, or you have problem executing one or more of them, feel free to ping me here. – user21436 Mar 17 '12 at 5:16
i am not getting it.Can you please explain a little more? Thanks so much. – vikiiii Mar 17 '12 at 5:17
To facilitate my writing out some details, it would help me if you can tell me about whether you have evaluated complex cube roots of unity in your class. – user21436 Mar 17 '12 at 5:18
Will it be equal to 1 ? Using the rule $$ w^3 =1 $$ where w is a complex root. – vikiiii Mar 17 '12 at 5:19
Well, an equation of degree $3$ has three roots, so what are the other two roots? So, one real root is $1$, there are two complex roots. Have you done some work on these in a class? – user21436 Mar 17 '12 at 5:21

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