In $\triangle ABC$, $X$ and $Y$ are points on the sides $AC$ and $BC$ respectively. If $Z$ is on the segment $XY$ such that $\frac{AX}{XC}=\frac{CY}{YB}=\frac{XZ}{ZY}$. Prove that the area of $\triangle ABC$ is given by $\triangle ABC = \left((\triangle AXZ)^{1/3}+(\triangle BYZ)^{1/3}\right)^3$ or alternatively $(\triangle ABC)^{1/3}=(\triangle AXZ)^{1/3}+(\triangle BYZ)^{1/3}$
I tried to use Menelaus theorem a bunch of times to leverage the ratios given but it does not get me anywhere useful so that I can relate them with the areas. I would be grateful if anybody could provide a solution. And preferably one that uses simple methods like similarity, Menelaus theorem. Ceva's theorem, etc., since this question is taken from an exercise on the above-mentioned topics. Of course, other solutions are always welcome, since it is always better to know multiple ways to solve the same problem.
Let the ratio $AX\over XC$ be $r$.
Then $S_{ACY}={r\over 1+r}S_{ABC}$, $S_{AXY}={r\over 1+r}S_{ACY}$, $S_{AXZ}={r\over 1+r}S_{AXY}$
$$S_{AXZ} = ({r\over 1+r})^3S_{ABC}$$
Similarly, we can find
$$S_{BYZ} = ({1\over 1+r})^3S_{ABC}$$
The equality thus follows.