Solve the following system of equations:

$$
\begin{aligned}
a x+(a-b) y+(a-c) z & =a^{2}+(b-c)^{2} \\
(b-a) x+\quad b y+(b-c) z & =b^{2}+(c-a)^{2} \\
(c-a) x+(c-b) y+\quad c z & =c^{2}+(a-b)^{2}
\end{aligned}
$$
Express $x$ and $y$ in terms of $a, b, c$, and $z$ from (1) and (2). Substituting the resulting expressions for $x$ and $y$ into (3), we solve the resulting equation for $z$. Multiplying (1) by $b$ and (2) by $(a-b)$, and then subtracting, $y$ is eliminated:

$$
\begin{gathered}
{\left[a b+(a-b)^{2}\right] x=b\left[a^{2}+(b-c)^{2}\right]+(b-a)\left[b^{2}+(c-a)^{2}\right]+} \\
+\left(2 b c-a c-b^{2}\right) z
\end{gathered}
$$

Multiplying (1) by $(a-b)$ and (2) by $a$, and then adding, $x$ is eliminated:

$$
\begin{aligned}
& {\left[(a-b)^{2}+a b\right] y=(a-b)\left[a^{2}+(b-c)^{2}\right]+a\left[b^{2}+(c-a)^{2}\right]+} \\
&+\left(2 a c-a^{2}-b c\right) z
\end{aligned}
$$

Now multiply (3) by $\left[a b+(a-b)^{2}\right]=\left(a^{2}-a b+b^{2}\right)$ and substitute the right-hand sides of (4) and (5) into the first two products on the left. After rearranging and simplifying the resulting first-degree equation in $z$, the coefficient of $z$ on the left is

$$
\begin{gathered}
E=(c-a)\left(2 b c-a c-b^{2}\right)+(c-b)\left(2 a c-a^{2}-b c\right)+ \\
+c\left[a b+(a-b)^{2}\right]=a^{2} b+a^{2} c+b^{2} a+b^{2} c+c^{2} a+c^{2} b-5 a b c
\end{gathered}
$$

We collect the terms on the right that do not contain $z$. Notice that the right-hand sides of (1)-(3), after expansion, can be written as

$$
S-2 b c, \quad S-2 a c, \quad S-2 a b
$$

with $S=a^{2}+b^{2}+c^{2}$. Therefore, the coefficient of $S$ is:

$$
\begin{gathered}
{\left[a b+(a-b)^{2}\right]-(c-a)[b+(b-a)]-(c-b)[(a-b)+a]=} \\
=3 a b-a c-b c
\end{gathered}
$$

We rearrange the remaining polynomial in decreasing powers of $c$:

$$
\begin{aligned}
& T=-2 a b\left[a b+(a-b)^{2}\right]+2 b c[b(c-a)+(a-b)(c-b)]+ \\
& +2 a c[(b-a)(c-a)+a(c-b)]= \\
& =4 a b c^{2}+\left(2 a^{3}-4 a^{2} b-4 a b^{2}+2 b^{3}\right) c-2 a b\left(a^{2}-a b+b^{2}\right)
\end{aligned}
$$

Thus, the entire right-hand side, rearranged in powers of $c$, is

$$
\begin{aligned}
& J=S(3 a b-a c-b c)+T=-(a+b) c^{3}+7 a b c^{2}+ \\
& +\left(a^{3}-5 a^{2} b-5 a b^{2}+b^{3}\right) c+a b\left(a^{2}+2 a b+b^{2}\right)
\end{aligned}
$$

Thus, from the equation $E z=J$, we have $z=J / E$. To divide the two polynomials, we rearrange $E$ in powers of $c$:

$$
\begin{gathered}
E=(a+b) c^{2}+\left(a^{2}-5 a b+b^{2}\right) c+a b(a+b), \quad \text { so } \\
z=-c+a+b
\end{gathered}
$$

Substituting this expression for $z$ into (4) and (5), we obtain first-degree equations for $x$ and $y$. It is convenient to rearrange the right-hand side of (4) in powers of $a$:

$$
\begin{gathered}
{[b+(b-a)] S-2 b^{2} c-2 a c(b-a)+\left(2 b c-a c-b^{2}\right)(a+b-c)=} \\
=-a^{3}+(2 b+c) a^{2}-\left(2 b^{2}+b c\right) a+\left(b^{3}+b^{2} c\right)
\end{gathered}
$$

Dividing this by $a b+(a-b)^{2}=a^{2}-a b+b^{2}$, we get

$$
x=-a+b+c
$$

and similarly

$$
y=-b+a+c
$$

Thus, if none of the expressions we divided by vanish, the solution to the system is:

$$
x=b+c-a, \quad y=c+a-b, \quad z=a+b-c
$$

Substitution shows that these expressions satisfy the system in all cases.

We provide a sufficient condition on the triple $(a, b, c)$ such that (7) is the unique solution to the given system.

The divisor in determining $x$ and $y$ is:

$$
a^{2}-a b+b^{2}=\left(a-\frac{1}{2} b\right)^{2}+\frac{3}{4} b^{2}
$$

which is positive unless $b=0$ and $a=b / 2=0$. In the latter case, the system becomes indeterminate. (From it, $z=-c$, $x+y=c$ follows if $c \neq 0$, and if $c$ is also 0, then we have three $0=0$ equations, which do not contain the unknowns and are thus satisfied by any triple $(x, y, z)$.) A similar situation arises if any other two of $a, b, c$ are 0. If, however, at most one of the three parameters is 0, then $x$ and $y$ are uniquely determined, provided $z$ is unique, which holds if the expression $E$ under (6) is not 0.

Without finding all triples $(a, b, c)$ for which $E \neq 0$, we show that this is always the case if $a, b, c$ are non-negative and at most one of them is 0. Indeed,

$$
\begin{gathered}
E=a^{2} b+a^{2} c+a b^{2}+a c^{2}+b^{2} c+b c^{2}-5 a b c= \\
=a^{2} c-2 a b c+b^{2} c+a b^{2}-2 a b c+a c^{2}+a^{2} b-2 a b c+b c^{2}+a b c= \\
=(a-b)^{2} c+(b-c)^{2} a+(c-a)^{2} b+a b c
\end{gathered}
$$

Here, no term is negative if $a, b, c$ are non-negative, and if all three numbers are positive, the last term is positive. If, for example, $c=0$ but $a \neq 0, b \neq 0$, then the second and third terms are positive.

From the work of János Csirik (Orosháza, M. Táncsics III. o. t.), with additions.