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In the name of God
If $x^2 + x+ 1 = 0$ find the value of $x^{1401} + x^{-1401}$.

I tried to factorize $x^{1401} + x^{-1401}$ by replacing $x$ with $a$ and $x^{-1}$ with $b$.

To factorize $a^{1401} + b^{1401}$, we can write it like this:

$$\begin{aligned} a^{1401} + b^{1401} &= (a^{467})^3 + (b^{467})^3\\ &= (a^{467} + b^{467})(a^{934} -a^{467}b^{467} + b^{934})\\ &=(a+b)(a^{466}-a^{465}b + a^{464}b^2 \\ &\quad-\cdots + b^{466})(a^{934} -a^{467}b^{467} + b^{934}) \\ &=(a+b)(a^{466}-a^{465}b + a^{464}b^2 -\cdots\\ &\quad+ b^{466})(a^2-ab+b^2)(a^{932} + a^{931}b \\ &\quad-a^{929}b^3 - \cdots + b^{932})\end{aligned}$$

But it doesn't really seem to help to solve the question.

I don't know much about more advanced math topics, but $x$ is certainly a complex number. How is it possible that $x^{1401} + x^{-1401}$ is a real one? Do 'i's get omitted?

Thanks for you help!

madfd adfd
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2 Answers2

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$\textbf{Hint:}$

$$x(x^2+x+1) = x^3+x^2+x = 0 \require{cancel}\implies x^3+\cancelto{0}{(x^2+x+1)} = 1$$

Ninad Munshi
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  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Jul 31 '23 at 14:30
  • Thanks for the answer. I have a question. does $x^{1401} + x^{-1401} -2 $ have factor $x^2 + x+ 1$ ? – madfd adfd Jul 31 '23 at 14:35
  • @madfdadfd Multiplying if by $x^{1401}$ we get a polynoomial $f(x)$ that is $\equiv 0\pmod{x^3-1}$ so $f(x)$ is divisibile by $x^3-1$ so also divisibile by its factor $x^2+x+1\ \ $ – Bill Dubuque Jul 31 '23 at 18:45
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$$x^3-1=(x-1)(x^2+x+1)=0$$ Therefore $$x^{1401}+x^{-1401}=(x^3)^{467}+(x^3)^{-467}=2$$

user
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