Solve the following system of equations:

$$
\begin{gathered}
a^{3} x+a y+z=a^{2} \ldots \\
x+y+z=1 \ldots \\
8 x+2 y+z=4 \ldots
\end{gathered}
$$

What cases arise when $a=1, a=2, a=-3$?
By subtracting the terms of equation 2) from the corresponding terms of equation 3), we get:

$$
7 x+y=3 \ldots
$$

Subtracting the terms of equation 2) from the terms of equation 1): $\left(a^{3}-1\right) x+(a-1) y=a^{2}-1$. If $a \neq 1$, then we can simplify by the factor $(a-1)$ and thus

$$
\left(a^{2}+a+1\right) x+y=a+1 \ldots
$$

Now from equations 4) and 5):

$$
\left(a^{2}+a-6\right) x=a-2 \ldots
$$

However,

$$
\begin{aligned}
& \qquad a^{2}+a-6=(a+3)(a-2) \\
& \text { if } \quad a \neq 2, \quad \text { then } \quad(a+3) x=1 \\
& \text { and if } \quad a \neq-3, \quad \text { then } \quad x=\frac{1}{a+3}
\end{aligned}
$$

According to equation 4)

$$
y=3-7 x=3-\frac{7}{a+3}=\frac{3 a+2}{a+3} \text {. }
$$

From equation 2)

$$
z=1-x-y=1-\frac{1}{a+3}-\frac{3 a+2}{a+3}=-\frac{2 a}{a+3}
$$

We assumed that $a$ is neither 1, nor 2, nor 3.

If $a=1$, then equations 1) and 2) are identical.

For $a=2$, equations 1) and 3) are identical.

In both cases, we will have two equations for calculating the three unknowns:

$$
x+y+z=1 \quad \text { and } \quad 8 x+2 y+z=4 \text {. }
$$

From here, $y=3-7 x, z=6 x-2$, which means we are faced with an indeterminate case. (There are infinitely many solutions!)

If $a=-3$ then equation 6) contains a contradiction; there is no finite value system that satisfies all three equations. ( $x, y, z$ - as their expressions show - become infinite!)

Ede Róka (Ref. g. V. o. Bp.)