$\begin{cases} x^3-y^3=19(x-y) \\ x^3+y^3=7(x+y) \end{cases}$ 
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?
 A: 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 $(x+y)^2, (x-y)^2$
A: After considering of cases $x=y$ or $x=-y$ use
$$7(x^2+xy+y^2)=19(x^2-xy+y^2).$$
A: 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 $y \neq \pm x.$

A: To avoid all possible problems, add the two equations to get
$$x^3-13 x+6 y=0 \implies y=\frac{1}{6} \left(13 x-x^3\right)$$ Plug in the first equation to end with
$$x \left(x^8-39 x^6+507 x^4-2665 x^2+4788\right)=0$$ So $x=0$ if a root.
For the remaining, let $z=x^2$ to make
$$z^4-39z^3+507z^2-2665z+4788=0$$ By inspection, $z=4$ and $z=7$ are solutions. Now, long division
$$\frac{z^4-39z^3+507z^2-2665z+4788 } {(z-4)(z-7) }=z^2-28 z+171=(z-19) (z-9)$$
