Let quadratic equation has roots $x_1,x_2$ and cubic polynomial has roots $x_1,y_1,y_2$ $x_1x_2$ and $x_1y_1y_2$ are rational, so $y_1y_2=kx_2$ for some rational $k$ $x_1+x_2$ and $x_1+y_1+y_2$ are rational, so $y_1+y_2-x_2=m$ for some rational $m$ $x_1y_1+x_1y_2+y_1y_2=x_1(m+x_2)+kx_2= k(x_1+x_2)+x_1x_2+ (m-k)x_1$ is rational, so $m=k$ But then $y_1y_2=kx_2,y_1+y_2=k+x_2 \to y_1 =k,y_2=x_2$ or $y_1=x_2,y_2=k \to $ one of roots is rational $\to$ contradiction