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

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

• Thank you for the response! $2x^3=14x$ and I think you missed $x=0$. – Katherine Jan 7 '20 at 20:15
• I left that for you in the second one:). Actually I missed it and have edited it in. – Ross Millikan Jan 7 '20 at 20:27

After considering of cases $$x=y$$ or $$x=-y$$ use $$7(x^2+xy+y^2)=19(x^2-xy+y^2).$$

• Can down-voter explain us, why did you do it? – Michael Rozenberg Jan 20 '20 at 5:31

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 $$y \neq \pm x.$$

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