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 $\gcd(x,y)=\gcd(a,x)=\gcd(a,y)=\gcd(a,b)=1$
 A: 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 $xy=0$.
A: Solving the system with respect to $a$ and $b$, we get:
$a=(x+y)^{1/3}$,
$b=(x^{2}-x y+y^{2})^{1/3}$;
multiplying the values of $a$ and $b$ and raising to the cube, we have:
$(a b)^{3}=x^{3}+y^{3}$;
hence the conclusion that the system of diophantine equations has no solutions.
A: $
\begin{cases} 
1)&x = a^3 - y   \\
2)&xy = \frac{a^6 - b^3}{3}   \\
\end {cases}
$
First, subsitute $1)$ into $2)$
$(a^3 - y)y = \frac{a^6 - b^3}{3}$
$3y\left(a^3-y\right)=a^6-b^3$
$3a^3y-3y^2=a^6-b^3$
$-3y^2+3a^3y-a^6+b^3=0$
$y = \frac{-3a^3\pm \sqrt{-3a^6+12b^3}}{2\left(-3\right)}$
$y = \frac{3a^3\pm\sqrt{12b^3-3a^6}}{6}$
Substitute  $y = \frac{3a^3\pm\sqrt{12b^3-3a^6}}{6}$ into $1)$
$x = \frac{3a^3\pm\sqrt{12b^3-3a^6}}{6}$
Subsitute $x = \frac{3a^3+\sqrt{12b^3-3a^6}}{6}$
and  $y = \frac{3a^3\pm\sqrt{12b^3-3a^6}}{6}$  into $1)$
$\frac{3a^3\pm\sqrt{12b^3-3a^6}}{6} +  \frac{3a^3\pm\sqrt{12b^3-3a^6}}{6}  = a^3$
$\:\frac{3a^3\pm \sqrt{12b^3-3a^6}}{3}\:\:=\:a^3$
$12b^3 = 3a^6 \implies 4b^3 = a^6$
$GCD(a,b) = 1$, Thus there are only trivial solutions $(a = 0, b = 0)$.
