Solve the following system of linear equations:

$$
\begin{array}{r}
x+y+z=1 \\
a x+b x+c z=h \\
a^{2} x+b^{2} y+c^{2} z=h^{2}
\end{array}
$$
$$
\begin{array}{r}
x+y+z=1 \\
a x+b x+c z=h \\
a^{2} x+b^{2} y+c^{2} z=h^{2}
\end{array}
$$

Multiply the first equation by $a$ and subtract it from the second, and then multiply it by $a^{2}$ and subtract it from the third. Then we get the following two equations:

$$
\begin{array}{r}
(b-a) y+(c-a) z=h-a, \\
\left(b^{2}-a^{2}\right) y+\left(c^{2}-a^{2}\right) z=h^{2}-a^{2} .
\end{array}
$$

Multiply the first by $(c+a)$ and subtract it from the second. Then we get

$$
\begin{gathered}
(b-a)(b+a-c-a) y=(h-a)(h+a-c-a) \\
(b-a)(b-c) y=(h-a)(h-c) \\
y=\frac{(h-a)(h-c)}{(b-a)(b-c)} .
\end{gathered}
$$

If we multiply (4) by $(b+a)$ and subtract it from (5), then

$$
\begin{gathered}
(c-a)(c+a-b-a) z=(h-a)(h+a-b-a) \\
(c-a)(c-b) z=(h-a)(h-b) \\
z=\frac{(h-a)(h-b)}{(c-a)(c-b)} .
\end{gathered}
$$

Finally,

$$
x=\frac{(h-b)(h-c)}{(a-b)(a-c)}
$$

(Jorga Gergely, 8th grade student, Arad).