Trying to solve the following problem:
Given a positive number $a$. It is known that equation $x^3+1=ax$ has exactly two positive roots, and the ratio of the larger one to the smaller one is $2018$, that is $x_1 / x_2 = 2018, \; x_1 > x_2$. Equation $x^3+1=ax^2$ also has exactly two positive roots. We need to prove that the ratio of the larger of them to the smaller is also equal to $2018$.
My idea is that we can represent left side of the equation as a difference of cubes: $x^3+1=(x+1)(x^2-x+1)$ and then we can try to add two equations, get the equation: $(x+1)(x^2-a(x+1)-1)=0$. It can be seen that one root is exactly negative. Can the roots of the second factor be connected with the relation of the roots of both equations?
Is it in that direction I'm moving or is there an easier way to solve the problem?
Hint: what happens if you put $x=\frac 1y$ in the first equation