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I have a task: Prove that if for each complex number u,w,z we have:

$uwz=1$ and $u+w+z=u^{-1}+w^{-1}+z^{-1}$

then at least one of them is equal 1.

I tried substituting $u=(uw)^{-1}$ to the second equality and prove that by denying, but I can't solve it.

Thanks in advance.

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$$ w = {1\over zu} \implies z+u+{1\over zu} = {1\over z} + {1\over u} +zu$$

so $$ z^2u+u^2z+1= u+z+z^2u^2$$

so $$z^2u(u-1)-z(u^2-1)+u-1=0$$

so $$(u-1)(z^2u-z(u+1)+1)=0$$

so if $u =1$ we are done else:

$$z^2u-zu-z+1=0\implies zu(z-1)-(z-1)=0$$

so if $z=1$ we are done else $zu = 1$ so $w=1$.

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If $uwz=1$ then $u^{-1}+w^{-1}+z^{-1}=uw+uz+wz$, and $$(X-u)(X-w)(X-z)=X^3-(u+w+z)X^2+(uw+uz+wz)X-uwz=\cdots$$ etc.

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Let $S=u+w+z$As observed in another answer, the equalities $uvw=1$ and $S=u^{+1}+w^{-1}+w^{-1}$ imply $uw+wz+zu=S$.

Now let's use Vieta's relations: $u,w$ and $z$ are the roots of the cubic polynomial $$Z^3-SZ^2+SZ-1=Z^3-1-SZ(Z-1)=(Z-1)(Z^2+Z+1-SZ),$$ which has $Z=1$ as a root.

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