Solve the following system of equations:

$$
\begin{gathered}
x+y+z=\frac{a b}{c}+\frac{b c}{a}+\frac{c a}{b} \\
c x+a y+b z=a b+b c+c a \\
c^{2} x+a^{2} y+b^{2} z=3 a b c
\end{gathered}
$$
First solution. The given system of equations can also be written as:

$$
\begin{gathered}
a b(c x-a b)+b c(a y-b c)+c a(b z-c a)=0 \\
c x-a b+a y-b c+b z-c a=0 \\
c(c x-a b)+a(a y-b c)+b(b z-c a)=0
\end{gathered}
$$

Let

$$
c x-a b=s, \quad a y-b c=u, \quad b z-c a=v
$$

then

$$
\begin{gathered}
a b s+b c u+c a v=0 \\
s+u+v=0 \\
c s+a u+b v=0
\end{gathered}
$$

from which \( s=0, u=0, v=0 \), so

$$
x=\frac{a b}{c}, \quad y=\frac{b c}{a}, \quad z=\frac{c a}{b}
$$

(Nándor Bayer, Losonc.)

Second solution 1 The determinant of the system of equations is:

$$
D=\left|\begin{array}{ccc}
1 & 1 & 1 \\
c & a & b \\
c^{2} & a^{2} & b^{2}
\end{array}\right|=\left|\begin{array}{ccc}
0 & 0 & 1 \\
c-b & a-b & b \\
c^{2}-b^{2} & a^{2}-b^{2} & b^{2}
\end{array}\right|=(a-b)(a-c)(c-b)
$$

Furthermore,

$$
\begin{aligned}
& D_{1}=\left|\begin{array}{ccc}
\frac{a b}{c}+\frac{b c}{a}+\frac{c a}{b} & 1 & 1 \\
a b+b c+c a & a & b \\
a b c+a b c+a b c & a^{2} & b^{2}
\end{array}\right|= \\
& =\frac{1}{a b c}\left|\begin{array}{ccc}
a^{2} b^{2} \cdot 1+b^{2} c^{2} \cdot 1+c^{2} b^{2} \cdot 1 & 1 & 1 \\
a^{2} b^{2} \cdot c+b^{2} c^{2} \cdot a+c^{2} a^{2} \cdot b & a & b \\
a^{2} b^{2} \cdot c^{2}+b^{2} c^{2} \cdot a^{2}+c^{2} a^{2} \cdot b^{2} & a^{2} & b^{2}
\end{array}\right|= \\
& \frac{a b}{c}=\left|\begin{array}{ccc}
1 & 1 & 1 \\
c & a & b \\
c^{2} & a^{2} & b^{2}
\end{array}\right|=\frac{a b}{c}(a-b)(a-c)(c-b)=\frac{a b}{c} D
\end{aligned}
$$

Similarly,

$$
D_{2}=\frac{b c}{a} D \quad \text { and } \quad D_{3}=\frac{c a}{b} D
$$

Therefore,

$$
x=\frac{D_{1}}{D}=\frac{a b}{c}, \quad y=\frac{D_{2}}{D}=\frac{b c}{a}, \quad \text { and } \quad z=\frac{D_{3}}{D}=\frac{c a}{b} \text {. }
$$

(Pál Sárközy, Pannonhalma.)[^0]


[^0]:    ${ }^{1}$ L.K.M.L.IX.175. Antal M.: On Determinants.