<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] Another obvious solution is $x=-1$. So we can write the equation as, $$(x^2-1)(78x^4+ax^3+bx^2+cx+78)=0$$ and comparing coefficients, we can easily find $a=c=-133$, $b=78$.
Now divide this (second polynomial) by $x^2$ and substitute $t=x+\frac1x$ to get a quadratic in $t$. It's easy to solve then.
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