Problem in solving equations

Question

While solving algebric I stuck on solving these pair of equations simulatneously: \begin{align} (x^2-y^2)(x-y)= 16xy\\ (x^4-y^4)(x^2-y^2)=640x^2y^2 \end{align} I have tried many substitution like $y=mx$ or $x+y=u$ and $x-y=v$ but nothing seems to work. Is there any other way out to solve these equations simultaneously?

Answer 1

Divide equation $2$ by equation $1$:

$$\frac{x^4-y^4}{x-y} = 40xy$$ $$\Rightarrow \frac{(x^2+y^2)(x^2-y^2)}{x-y} = 40xy$$ $$\Rightarrow (x^2+y^2)(x+y)= 40xy$$ $$\Rightarrow x^3+x^2y+y^2x+y^3= 40xy \tag{3}$$

And now expand equation $1$: $$x^3-x^2y-y^2x+y^3 = 16xy $$ $$\Rightarrow x^3+y^3 = 28xy \tag{4}$$

therefore:

$$x^2y+y^2x = 12xy$$ $$xy(x+y) = 12xy$$ $$x+y = 12 \tag{5}$$

Substituting this into the first equation gives a quadratic, which results in $x=3,9$, and $y=9,3$; so the solutions are $(3,9)$ and $(9,3)$.

We also have the case where $x-y=0$ or $x=y$. This gives the solution $(0,0)$ as well.

Answer 2

The resultants of the two polynomials (when written as $f(x,y)=0$) with respect to both variables are, up to a constant, $x^{12}(x-3)(x-9)$ and $y^{12}(y-3)(y-9)$. This leaves us with only nine cases to check, and we find that there are three distinct solutions: $(x,y)=(0,0),(3,9),(9,3)$.