Problem 1314

Solution. Let the positive roots of the equation $x^{3}+1=a x$ be denoted by $x_{1}$ and $x_{2}$ $\left(0<x_{1}<x_{2}, x_{2}: x_{1}=2018\right)$. Substitute them into the equation and divide the two resulting equations by $x_{1}^{3}$ and $x_{2}^{3}$:

$$\begin{aligned} & x_{1}^{3}+1=a x_{1} \Longleftrightarrow 1+\left(\frac{1}{x_{1}}\right)^{3}=a\left(\frac{1}{x_{1}}\right)^{2} \\ & x_{2}^{3}+1=a x_{2} \Longleftrightarrow 1+\left(\frac{1}{x_{2}}\right)^{3}=a\left(\frac{1}{x_{2}}\right)^{2} \end{aligned}$$

From the formulas, it is clear that $\frac{1}{x_{1}}$ and $\frac{1}{x_{2}}$ are the positive roots of the equation $x^{3}+1=a x^{2}$. According to the condition, there are exactly two such roots, and we need to find the ratio of the larger to the smaller. It is evident that $\frac{1}{x_{2}}<\frac{1}{x_{1}}$. Therefore, $\frac{1}{x_{1}}: \frac{1}{x_{2}}=x_{2}: x_{1}=2018$.

## Criteria

2 6. The work notes that when the first equation from the condition is divided by $x^{3}$ and the substitution $\frac{1}{x} \rightarrow x$ is made, the first equation transforms into the second, but there is no further progress.

3 6. The work proves that the second equation has positive roots $\frac{1}{x_{1}}$ and $\frac{1}{x_{2}}$, but the ratio of the larger to the smaller is found incorrectly (for example, the larger and smaller are swapped).

4 6. Any complete and correct solution.