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the question

Show that if $x + y + z = xy + xz + yz$ then $x ^ 2 * (1 - y ^ 2) + y ^ 2 * (1 - z ^ 2) + z ^ 2 * ( 1 - x ^ 2) = 2(x + y + z)(xyz - 1)$

the idea

First of all i though of breaking all the products and i got to:

$x^2-x^2y^2+y^2-y^2z^2+z^2-z^2x^2=2x^2yz+2xy^2z+2xyz^2-2x-2y-2z$

Then i squared the given equality and got to

$x^2+y^2+z^2+2(xy+yz+xz)=x^2y^2+y^2z^2+x^2z^2+2x^2yz+2xy^2z+2xyz^2$

I tried processing it, but got to nothing usegul. Hope one of you can help me! Thank tou!

IONELA BUCIU
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  • Hint: try expanding $(x + y + z)^2 = (xy + xz + yz)^2$. – Theo Bendit Mar 04 '24 at 12:16
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    The two equations you wrote are literally the same. Just substitute $2(x+y+z)$ with $2(xy+yz+zx)$. You have (almost) the full solution. – Aig Mar 04 '24 at 12:24
  • Side note: Ionela, I've noticed that for several of your posts where you shown what you've tried, you're often a slight step away from completing your solution. If so, I encourage you to verbalize what the gap in your proof is / why you are stuck. EG What did you mean by "tried processing it"? $\quad$ For example, in this case, you wanted to prove that those 2 equations implied each other. So If we wrote them as $f(x) = 0 , g(x) = 0 $, then looked at when $f(x) = g(x)$, you might realize that $\Leftrightarrow 2(x+y+z) = 2(xy+yz+zx)$ like Aig pointed out. – Calvin Lin Mar 04 '24 at 16:24
  • Move negative terms. – Bob Dobbs Mar 04 '24 at 17:50

1 Answers1

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$$x^{2}\cdot \left(1-y^{2}\right)+y^{2}\cdot \left(1-z^{2}\right)+z^{2}\cdot \left(1-x^{2}\right)-2 \left(x+y+z\right) \left(x y z-1\right)=-\left(x y+x z+y z+x+y+z+2\right)\cdot \underbrace{\left(x y+x z+y z-(x+y+z)\right)}_0=0$$

gpmath
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