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Show that for any $x, y \in \mathbb R$ with $x + y \neq 0,xy\neq 0$

$$p(x,y) := x^6-2 x^5 y+2 x^5-x^4 y^2-2 x^4 y+x^4+4 x^3 y^3+2 x^3 y-x^2 y^4-4 x^2 y^3-4 x^2 y^2+2 x^2 y-2 x y^5+6 x y^4+2 x y^3+y^6-2 y^5-y^4-2 y^3+y^2 \neq 0$$

I'm sorry,I forget $xy\neq 0$,Now I think it's hold?

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    $\begingroup$ It isn't even 0 when $x+y=0$ $\endgroup$
    – Yifan Zhu
    Commented Dec 9, 2017 at 10:48
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    $\begingroup$ This question would be much improved if you included some context and/or motivation. Give people a reason to care about this problem. $\endgroup$
    – Todd Trimble
    Commented Dec 9, 2017 at 15:29
  • $\begingroup$ It is from a elementary queston: if $$\left(\sqrt{y^{2} - x\,\,}\, - x\right)\left(\sqrt{x^{2} + y\,\,}\, - y\right)=y$$ then $x+y=0$. $\endgroup$
    – math110
    Commented Dec 9, 2017 at 15:34
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    $\begingroup$ Then that should be included in the question. Not in a comment. $\endgroup$
    – Todd Trimble
    Commented Dec 9, 2017 at 15:38
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    $\begingroup$ Meanwhile, however, I tried seeing what solutions there are when $y = 1$. I seem to get $p(x, 1) = x^6 - 2x^4 + 6x^3 - 7x^2 + 6x - 3$. (Even if I made an arithmetic mistake, the leading term is $x^6$ and the constant term is $-3$.) At $x = 0$ we have $p(0, 1) = -3$, and $p(x, 1) \to \infty$ as $x \to \infty$. By the intermediate value theorem, there is $r > 0$ such that $p(r, 1) = 0$. And then $(x, y) = (r, 1)$ satisfies $xy \neq 0$ and $x + y \neq 0$. $\endgroup$
    – Todd Trimble
    Commented Dec 9, 2017 at 15:43

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I find p(-1,0)=0. Typo?

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