Originally Posted by

**rochosh** Solving the first equation$\displaystyle (1)$ further enough you'll get:

$\displaystyle (x^3+y^3)^2=8x^3y^3...........(I)$

Factorizing and reducing the second one$\displaystyle (2)$:

$\displaystyle (x^3-y^3)^3(x^3+y^3)^2=(6x^4y^4)^3..........(II)$

Replacing $\displaystyle (I)$ in $\displaystyle (II)$:

$\displaystyle (x^3-y^3)^{3}8x^3y^3=(6x^4y^4)^3$

Solving this final equation in order to find the required one above and you'll get

$\displaystyle -1=\frac{3x^3y^3}{y^3-x^3}$

Reorganizing my attempt to post it, I casually found out the solution.

In any case another way to solve this problem will be appreciated.

Thanks for the remarking.