Solve the following equation

$$
\left|\begin{array}{ccc}
a-x & c & b \\
c & b-x & a \\
b & a & c-x
\end{array}\right|=0
$$[^0]


[^0]:    ${ }^{1}$ L. K. M. L. IX. k. 175. 1. . On Determinants.
By adding the second and third rows to the first row of the determinant and then factoring out the common divisor of each term in the first row, we get:

$$
(a+b+c-x)\left|\begin{array}{ccc}
1 & 1 & 1 \\
c & b-x & a \\
b & a & c-x
\end{array}\right|=0
$$

By subtracting the first column from the others, we get:

$$
(a+b+c-x)\left|\begin{array}{ccc}
1 & 0 & 0 \\
c & b-c-x & a-c \\
b & a-b & c-b-x
\end{array}\right|=0
$$

or

$$
(a+b+c-x)\left[x^{2}-(b-c)^{2}-(a-b)(a-c)\right]=0
$$

From which

$$
\begin{gathered}
x_{1}=a+b+c \\
x_{2}=\sqrt{a^{2}+b^{2}+c^{2}-a b-a c-b c} \\
x_{3}=-\sqrt{a^{2}+b^{2}+c^{2}-a b-a c-b c}
\end{gathered}
$$

(György Schuster, Budapest.)

The problem was also solved by: Elemér Bauer, N. Bayer, A. Czúcz, N. Ehrenfeld, H. Fodor, J. Guman, P. Hajdu, M. Hermann, E. Kirchknopf, E. Kiss, R. Kürth, Gy. Meleghy, E. Mellinger, C. Neubauer, S. Pichler, L. Spitzer, P. Sárközy, Gy. Schwarz, S. Schwarz, Science Club, Bpest, VII. district.