What is the solution of $\frac{x^3}{y^3+z^3}+\frac{y^3}{x^3+z^3}+\frac{z^3}{y^3+x^3}=1$ in rational?

Question

I have tried to solve the following equation in rational($x,y,z$ are rational numbers) $$\frac{x^3}{y^3+z^3}+\frac{y^3}{x^3+z^3}+\frac{z^3}{y^3+x^3}=1$$ , I can't find such pairs of rational numbers $(x,y,z)$ for which that function have solution , I have assumed $x=0,y=1$, I have got a complex solutions for z(unit complex roots) , I think it must to add another identity for which we have solution,And also I used $(1/(x+y+z))^3=1$ to excpand some terms in the titled identity but no result ? Any help ?

Edit I have added this related link to this question which is provided by @Madara Uchiha in the comment

Note: The motivation of this question is to check whether we can write $1$ in sum of cubic rational terms ,And I suspect if such solutions exist they must be large number

Answer 1

some idea

If $(x,y,z) $ is a solution, then

$ (x^3,y^3,z^3) $ will necessarily be a rational solution of

$$\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=1$$

Answer 2

Here's a roadmap to a solution; the details are too tedious for me to figure out.

First note that if $(x,y,z)\in\Bbb{Q}^3$ satisfies $$\frac{x^3}{y^3+z^3}+\frac{y^3}{x^3+z^3}+\frac{z^3}{y^3+x^3}=1,\tag{1}$$ then so does $(kx,ky,kz)$ for every nonzero integer $k$, so it suffices to find the integral solutions to $(1)$.

Now suppose $(x,y,z)\in\Bbb{Z}$ is and integral solution to $(1)$. Then clearing denominators and rearranging a bit we see that $$x^9+y^9+z^9+x^3y^3z^3=0.$$ Then setting $(X,Y,Z):=(x^3,y^3,z^3)$ we get the following cubic equation in $X$, $Y$ and $Z$, $$X^3+Y^3+Z^3+XYZ=0,\tag{2}$$ and then using the answer to this question as suggested in the comments, we can transform this to an elliptic curve with short Weierstrass form $$Y^2=X^3+aX+b,$$ where $$a=\frac{3085015005}{614656}\qquad\text{ and }\qquad b=-\frac{55345728797073}{240945152}.$$ Then LMFDB tells me that the corresponding minimal Weierstrass equation is in fact $$y^2+xy+y=x^3-4x+6,\tag{3}$$ and that this elliptic curve has rank only the following five affine rational points: $$(1,-1),\quad(2,2),\quad(2,-5),\quad(9,23),\quad(9,-33).$$ Of course equation $(2)$ has the obvious affine solutions (i.e. with $Z=1$) $$(1,-1),\quad (-1,1),\quad (-1,-1),$$ that do not correspond to solutions to $(1)$, as they have $x^3+z^3=0$ or $y^3+z^3=0$.

Then there are two more rational solutions to $(2)$ with $Z=1$, and they can be made explicit by determining the rational transformation between $(2)$ and $(3)$. If these $X$- and $Y$-values of these two rational solutions are not perfect cubes, then the original equation $(1)$ has no rational solutions, which seems likely.