<s>[INST] A Diophantine equation: solve $(3x^2+y^2-4y-17)^3-(2x^2+2y^2-4y-6)^3=(x^2-y^2-11)^3$ (without using Fermat's last theorem) Solve this Diophantine equation: $(3x^2+y^2-4y-17)^3-(2x^2+2y^2-4y-6)^3=(x^2-y^2-11)^3$
My attempt (use Fermat's last theorem)
$$(3x^2+y^2-4y-17)^3-(2x^2+2y^2-4y-6)^3=(x^2-y^2-11)^3$$
$$\Leftrightarrow (3x^2+y^2-4y-17)^3=(x^2-y^2-11)^3+(2x^2+2y^2-4y-6)^3$$
Use Fermat's last theorem, we get:
$$
\left\{
\begin{array}{c}
x^2-y^2-11=0 \\ 
3x^2+y^2-4y-17=2x^2+2y^2-4y-6
\end{array}
\right.$$
or 
$$
\left\{
\begin{array}{c}
 2x^2+2y^2-4y-6=0\\ 
3x^2+y^2-4y-17=x^2-y^2-11
\end{array}
\right.$$
Then we continue...
My question is, is there another way to solve that other than Fermat's last theorem? I see that using Fermat's last theorem is like crack a nut by a sledgehammer.
 [/INST] Let
$$3x^2 + y^2 - 4y -17 = a\text{ ,}$$
$$-(2x^2 +2y^2 +4y -6)= b$$
and observe
$$x^2-y^2 -11= a+b$$
Your problem may be reexpressed as $$a^3 + b^3=(a+b)^3$$ implying
$$3ab(a+b)=0.$$ This gives your proposed solution set, plus the case where $a=0$.
</s>