Rational solutions of $x^3+y^3=9$
This is from Albert Beiler, Recreations In The Theory Of Numbers: The Queen Of Mathematics Entertains. Chapter 25 is called Tilts and Tourneys. This is problem 83, on page 303. The answer is on pages 335-336, and confirms that the desired quantity is rational and positive $x,y$ such that $x^3 + y^3 = 9,$ smallest common denominator, but $(x,y) \neq (2,1)$ and $(x,y) \neq (1,2).$ He gives the specific answer, the common denominator is quite large.
The doubling construction through the point $(2,1)$ gives the new rational point $$ \left( \frac{20}{7} , \; \;\frac{-17}{7} \right) $$
It is not necessary to have a cubic curve in any special format to perform doubling, maybe it is called duplication. It is just calculus. We begin with $x^3 + y^3 = 9.$ We implicitly find $3 x^2 + 3 y^2 y' = 0,$ or $$ y' = - \left( \frac{x}{y} \right)^2 $$ For example, at the rational point $(2,1)$ we find $y' = -4.$ Next we parametrize the tangent line through $(2,1)$ by variable $t$ as $$ x = 2 + t, \; \; y = 1 - 4t $$ We set $x^3 + y^3 = 9$ and arrive at $$ -63t^3 + 54 t^2 = 0.$$ That is, $$ 9 t^2 (6 - 7 t) = 0. $$ Therefore, for nonzero $t,$ we have $t = 6/7.$ then $$ x = 2 + (6/7) = 20/7, $$ $$ y = 1 - (24/7) = -17/7 $$
Joining to $(1,2)$ gives $$ \left( \frac{919}{438} , \; \;\frac{-271}{438} \right) $$
ADDENDUM: I worked out duplication for a rational point $(a,b)$ on a curve where we make $x^3 + y^3 = \mbox{constant}.$ The value of the constant does not enter; we get $$ (a,b) \mapsto \left( \frac{a(a^3 + 2b^3)}{a^3 - b^3}, \; \; \frac{-b(2 a^3 + b^3)}{a^3 - b^3} \right). $$ In particular, $$ \left( \frac{20}{7} , \; \;\frac{-17}{7} \right) \mapsto \left( \frac{-36520}{90391} , \; \;\frac{188479}{90391} \right) $$ is not in the first quadrant. $$ \left( \frac{919}{438} , \; \;\frac{-271}{438} \right) \mapsto \left( \frac{676702467503}{348671682660} , \; \;\frac{415280564497}{348671682660} \right). $$ This is the answer Beiler supplies.
I drew a picture. It seems quite believable that Beiler's answer really is the first rational point that returns to the first quadrant, after leaving it.
I rotated the diagram by $45^\circ$ with $u = x-y, \; v = x+y.$ This was for my convenience in drawing the picture using points from a calculator, as $v^3 + 3 u^2 v - 36 = 0$ and Cardano gives $$v = \sqrt[3]{ \sqrt{324 + u^6} + 18} - \sqrt[3]{\sqrt{324 + u^6} - 18}$$ Then I made another picture with more points, including Beiler's solution. Most of the lines were light green, I made the line that duplicates and produces that in light blue.
