<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] You can divide by variables if you ensure they are not zero.  Here, you can consider the cases $x=y$ and $x=-y$ first.  If $x=y$ the first equation is trivial and the second becomes $2x^3=14x$ or $x=0,\pm \sqrt 7$.  You can do the same for $x=-y$ and find a pair of solutions.  
Then decree that $x+y \neq 0, x-y \neq 0$ and divide away.  Once you do that, you can subtract the two equations to get $2xy=12$ and use that to get expressions for $(x+y)^2, (x-y)^2$
</s>