<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] 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
$$\frac{z^4-39z^3+507z^2-2665z+4788 } {(z-4)(z-7) }=z^2-28 z+171=(z-19) (z-9)$$
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