<s>[INST] $\begin{cases} x^3-y^3=19(x-y) \\ x^3+y^3=7(x+y) \end{cases}$ 
I should solve the following system: $$\begin{cases} x^3-y^3=19(x-y) \\
 x^3+y^3=7(x+y) \end{cases}$$
  by reducing the system to a system of second degree. 

We can factor:
 $$\begin{cases} (x-y)(x^2+xy+y^2)=19(x-y) \\
 (x+y)(x^2-xy+y^2)=7(x+y) \end{cases}$$
I really don't want to divide the equations by $x-y$ and $x+y$, respectively. I am taught to divide by expressions containing variables only in special cases. Is there any other way here?
 [/INST] The thing that catches my eye is that $7,19$ are primes congruent to $1 \pmod 3,$ therefore integrally represented by both $x^2 + xy+y^2, \; x^2 -xy + y^2,$ meaning there are integer points on the two ellipses. It is worth drawing both, by hand (a valuable skill), to see whether that gives a simplified answer when $y \neq \pm x.$

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