<s>[INST] Solve $x^5=\frac{133x-78}{133-78x}$ Solve $$x^5=\dfrac{133x-78}{133-78x}$$ We have $D:133-78x\ne0,x\ne\dfrac{133}{78}$. In $x\in D$ the given equation is equivalent to $$x^5(133-78x)=133x-78\\78x^6-133x^5+133x-78=0$$ As $x=0$ is not a solution, we can divide both sides by $x^3\ne0$ to get $$78x^3-133x^2+\dfrac{133}{x^2}-\dfrac{78}{x^3}=0\\78\left(x^3-\dfrac{1}{x^3}\right)-133\left(x^2-\dfrac{1}{x^2}\right)=0$$ I am not able to come up with a subtle substitution. It would have been easier if we had $x^2\color{red}{+}\dfrac{1}{x^2}.$
$x=1$ is an "obvious" solution. What else? Thank you!
 [/INST] Let $(x^3-\frac{1}{x^3})=\sum_{n=3}^3x^n-x^{-n}$
$(x^2-\frac{1}{x^2})=\sum_{n=2}^2x^n-x^{-n}$
Then solve:
\begin{equation}
\sum_{n=3}^3x^n-x^{-n}=\frac{133}{78}\sum_{n=2}^2x^n-x^{-n}
\end{equation}
Computationally, that gives $x=-1, x=2/3, x=1, x=3/2$. Plus complex solutions.
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