<s>[INST] 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 $\gcd(x,y)=\gcd(a,x)=\gcd(a,y)=\gcd(a,b)=1$
 [/INST] $
\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 $(a = 0, b = 0)$.
</s>