$$\frac{x^5-a}{x^2-y}=\frac{y^5-b}{y^2-x}=5(xy-1)$$ solve the equation for x,y;a,b are arbitary
here is a problem from Ramnujan's Notebook.The procedure is
let$$x=\alpha+\beta+\gamma=S_1,y=\alpha\beta+\beta\gamma+\gamma\alpha=S_2,\alpha\beta\gamma=S_3=1$$
$$S_1^5-a=5(S_1S_2-S_3)(S_1^2-S_2)$$ $$ S_2^5-b=5(S_1S_2-S_3)(S_2^2-S_1) $$ we set $S_1,S_2,S_3$ with the aim of letting notes more concise and making the connection to Newton's identical(Newton's identical had not appear in the original text.It's from my idea to simplify the caculation)
then the text gives out the result without specific caculation but only a piece of word"Comparing coefficients of like terms on each side of each equation, we deduce that"
$$ \begin{equation} a=\alpha^5+\beta^5+\gamma^5 \\ b=(\alpha\beta)^5+(\beta\gamma)^5+(\gamma\alpha)^5 \end{equation} $$ then,of course,the problem has been simplified into a very simple cubic equation by using Vieta's theorem
$$t^3-at^2+bt-1=0$$
$\alpha^5,\beta^5,\gamma^5$ are solution for this cubic equation
the most crucial issue is that i don't konw how to caculate the a,b from the equation by so called "comparing coefficients of like terms"?
and I ask AI,it give interpration about Newton's Identical.I think is not a general explain,because it can only solve the first equation about a,the second equation cannot be solved by it.
and another idea is noticing that $S_1^5$ will generate tenth degree terms $\alpha^5,\beta^5,\gamma^5$,then the RHS can only generate fifth degree terms,so a must contain tenth terms to let the equation holds.But it can explain why we can be sure of that the other degree terms' coefficient can match
Of course we can guess $a,b$ is a formed by these 5th degree terms,but I want to know the general approach to figure out this type of problem,and guessing cannot be a general method(
here is another problem form Notebook which can be solved by using the similiar idea
$$\frac{x^7-a}{(x^2-y)^2+x}=\frac{y^7-b}{(y^2-x)^2+y}=7(xy-1) $$ solve this for x,y
