Problem 637

Solution. Let's bring the desired expression to a common denominator: $\frac{y^{2} z^{2}+x^{2} z^{2}+x^{2} y^{2}}{x y z}$. The polynomial has 3 different real roots, since $\mathrm{P}(-100)0, \mathrm{P}(0)0$. By Vieta's theorem $x+y+z=2, x y+x z+y z=-9, x y z=1$.

$$\begin{aligned} & x^{2} y^{2}+x^{2} z^{2}+y^{2} z^{2}=(x y+x z+y z)^{2}-2\left(x^{2} y z+y^{2} x z+z^{2} x y\right) \\ & =(x y+x z+y z)^{2}-2 x y z(x+y+z)=81-2 * 1 * 2=77 \end{aligned}$$

Answer: 77.