the system of diophantine equations: $x+y=a^3$; $xy=\dfrac{a^6-b^3}{3}$ has only trivial solutions. Without using Fermat's Last Theorem, how can one prove that the following system of diophantine equations has only trivial solutions:
$$x+y=a^3$$
$$xy=\dfrac{a^6-b^3}{3}$$
We suppose of course that $\gcd(x,y)=\gcd(a,x)=\gcd(a,y)=\gcd(a,b)=1$

If $a$, $b$, $x$ and $y$ are integers with $\gcd(x,y)=\gcd(a,b)=1$ such that
$$x+y=a^3\qquad\text{ and }\qquad xy=\frac{a^6-b^3}{3},$$
then $b^3=a^6-3xy$ and $a^6=(x+y)^2=x^2+xy+y^2$, so
$$b^3=(x^2+xy+y^2)-3xy=x^2-2xy+y^2=(x-y)^2.$$
It follows that for some integer $c$ we have $b=c^2$ and $x-y=c^3$. Then
$$x=\frac{a^3+c^3}{2}\qquad\text{ and }\qquad y=\frac{a^3-c^3}{2},$$
from which it follows that
$$\frac{a^6-c^6}{4}=\frac{a^3+c^3}{2}\frac{a^3-c^3}{2}=xy=\frac{a^6-b^3}{3}=\frac{a^6-c^6}{3}.$$
This shows that $a^6=c^6$ and hence that $xy=0$.
