<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] $$(3x^2+y^2-4y-17)^3-(2x^2+2y^2-4y-6)^3=(x^2-y^2-11)^3$$
If we put $A=3x^2+y^2-4y-17$ and $B=2x^2+2y^2-4y-6$, observe that $A-B=x^2-y^2-11$ (the factor on the right). 
In other words, we have $a^3-b^3=(a-b)^3$
Expand this to get $a^3-b^3=a^3-3a^2b+3ab^2-b^3$
So $3ab(a-b)=0$ is the simplified equation. You can then show that $a=0, b=0$ or $a-b=0$ has no integer solution. 
</s>