An algebra problem by Richard Bao

$(\frac { 1 }{ x } +x=1)\ast { x }^{ 6 }\quad returns\quad { x }^{ 5 }+{ x }^{ 7 }={ x }^{ 6 }\\ { x }^{ 7 }={ x }^{ 6 }-{ x }^{ 5 }\\ now,\quad since\quad (\frac { 1 }{ x } +x=1)\ast { x }^{ 5 }\quad returns\quad { x }^{ 6 }={ x }^{ 5 }-{ x }^{ 4 },\\ { x }^{ 7 }=-{ x }^{ 4 }\\ following\quad this\quad method,\quad we\quad get\quad that\quad { x }^{ 4 }={ x }^{ 3 }-{ x }^{ 2 },\quad and\quad thus\\ { x }^{ 7 }={ -x }^{ 3 }+{ x }^{ 2 }\\ Again,\quad we\quad know\quad that\quad { x }^{ 3 }={ x }^{ 2 }-{ x }^{ 1 },\quad so\\ { x }^{ 7 }={ x }^{ 1 }\\ Therefore,\quad \frac { { 1 } }{ { x }^{ 7 } } +{ x }^{ 7 }=1$

Solving the equation gives $x=\dfrac{1+\sqrt{-3}}{2}=\omega$

where $\omega$ is the imaginary sixth root of unity.

So $\omega^6=1$

$\omega^7=\omega×\omega^6=\omega$

Or $\omega^7+\dfrac{1}{\omega^7}=\omega+\dfrac{1}{\omega}=1$

x^2 - x + 1 = 0

x = 1/ 2 + j Sqrt (3)/ 2 OR 1/ 2 - j Sqrt (3)/ 2

Since | x | = 1, only angle but not magnitude need to be considered.

Direct substitution need no binomial expansion.

Take either one of them example x = 1/ 2 + j Sqrt (3)/ 2,

x^7 = 1/ 2 - j Sqrt (3)/ 2 and (1/ x)^7 = (Conjugate x)^7 = 1/ 2 + j Sqrt (3)/ 2;

1/ x^7 + x^7 = 1