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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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    If these hints are not sufficient, or you have problem executing one or more of them, feel free to ping me here.2012-03-17
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    i am not getting it.Can you please explain a little more? Thanks so much.2012-03-17
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    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.2012-03-17
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    Will it be equal to 1 ? Using the rule $$ w^3 =1 $$ where w is a complex root.2012-03-17
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    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?2012-03-17
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    @Kannappan Sampath: it is not necessary "Recall Euler's Form of a complex number and De-Moivre's Theorem". The answer follows immediately if one substitutes $-1$ for $x^3$ in the second equation.2012-03-17
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    @miracle173 It's quite possible as I did not work out the details, but you'll still be let with $-x-\frac 1 x$ and then, you can take conjugate and multiply. Maybe, I'll add this in if this is what you're saying.2012-03-17
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    @Kannappan Sampath: but $-x-1/x=-(x+1/x)$ an d the value of $x+1/x$ is known by the initial equation2012-03-17
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    @miracle173 Yeah, I am sorry. I did not take a look. So, the solution gets even more slick. Thanks for pointing it out.2012-03-17
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    @miracle173 I have edited to add your point. How does it look? (Thanks for telling anyway.)2012-03-17