For a positive rational $\frac{m}{n}$, find another positive rational $\frac{s}{t}$ such that $\frac{t^3m}{s^3n}=\frac{p}{q}$ lies between $\frac12$ and $2$. Such a rational clearly exists. Then $\frac{p}{q}=\frac{(p+q)^3+(2p-q)^3}{(p+q)^3+(2q-p)^3}$, and all four cubes are positive. Then just multiply back, giving $\frac{m}{n}=\frac{(s(p+q))^3+(s(2p-q))^3}{(t(p+q))^3+(t(2q-p))^3}$.