The roots of the equation $x^{4}-16 x^{3}+86 x^{2}-176 x+105=0$ form an arithmetic progression, where the eighth and thirteenth terms $\left[a_{8}=r\right.$ and $\left.a_{13}=R\right]$ are equal to the radii of two spheres; if the smaller sphere is shot through the center of the larger one, what will be the volume of the perforated larger sphere, and what is the surface area of the hole?
The given equation can also be written as:

$$
\begin{gathered}
x^{4}-16 x^{3}+86 x^{2}-176 x+105=x^{2}\left(x^{2}-4 x+3\right)-12 x^{3}+83 x^{2}-176 x+105= \\
=x^{2}\left(x^{2}-4 x+3\right)-12 x\left(x^{2}-4 x+3\right)+35\left(x^{2}-4 x+3\right)= \\
=\left(x^{2}-4 x+3\right)\left(x^{2}-12 x+35\right)=(x-1)(x-3)(x-5)(x-7)=0
\end{gathered}
$$

Thus, the roots of the given equation are:

$$
x_{1}=1, x_{2}=3, x_{3}=5, x_{4}=7
$$

The roots of the equation therefore form an arithmetic sequence of odd numbers, and thus

$$
a_{8}=r=15 \text { and } A_{13}=R=25
$$

The volume of the drilled sphere (K. M. L. VI. 528. problem):

$$
v=\frac{4 \pi}{3} \sqrt{\left(R^{2}-r^{2}\right)^{3}}=\frac{\pi}{3} \cdot 32000
$$

The surface area of the hole:

$$
f=4 \pi r \sqrt{R^{2}-r^{2}}=1200 \cdot \pi
$$

(Béla Bayer, Losoncz.)

The problem was also solved by: J. Baumann, K. Czank, F. Faith, J. Filkorn, K. Freudenberg, I. Hein, M. Holzmann, T. Kerekes, B. Krausz, Gy. Krisztián, Gy. Perl, Gy. Póka, G. Sasvári, J. Sasvári, J. Scharff, A. Singer, K. Smodics, S. Wohlstein.