2016/054) Let $a,b,c$ be rational and one of the roots of $ax^3+bx+c=0$ is equal to product of other two roots. Prove that this root is rational.

we can devide by a to get $x^3+dx+e=0$ where $d=\frac{b}{a},d=\frac{c}{a}$ $d,e$ are rational.
Let $\alpha,\beta,\alpha\beta$ be the three roots
so we get using vieta's relations
$\alpha+\beta+\alpha\beta= 0\cdots(1)$
$\alpha\beta+\alpha \alpha\beta + \beta\alpha\beta  = d\cdots(2)$
$\alpha\beta\alpha\beta= \alpha^2\beta^2= -e\cdots(3)$
from (2)
$\alpha\beta+ \alpha\beta(\alpha+\beta)  = d$
or $\alpha\beta+ \alpha\beta(\alpha+\beta)  = d$
or $\alpha\beta+ \alpha\beta(-\alpha\beta)  = d$ (Using (1)
or $\alpha\beta - (\alpha\beta)^2  = d$
or $\alpha\beta + e  = d$
or $\alpha\beta = d - e$
hence the root  $\alpha\beta$ is rational