By RRT, that one rational root will be either $1$ or $-1$. Hence, either $(x+1)$ or $(x-1)$ divides the equation. Each of them gives a value of $c$. Then it's easy to find the roots of a quadratic

Pinko wrote: We are given the equation $x^3-cx^2+(c-3)x+1=0$, where $c$ is an arbitrary number. Prove that, if the equation has at least one rational root, then all of its roots are rational. $ m- $ rational $ m^3-cm^2+(c-3)m+1=0 $ $ c=\frac{m^3-3m+1}{m^2-m} $