$\begin{cases} x^3-y^3=19(x-y) \\ x^3+y^3=7(x+y) \end{cases}$ 
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?

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 $(x+y)^2, (x-y)^2$
