For what value of the real parameter $v$ does the system of equations

$$
\begin{aligned}
x+y+z & =v \\
x+v y+z & =v \\
x+y+v^{2} z & =v^{2}
\end{aligned}
$$

have no solution?
Solution. If $v=0$, then from equation (2) we get $x=-z$, and from equation (3) we get $x=-y$, which is only possible if $x=y=z=0$. We can assume that $v \neq 0$.

Subtract (1) from (2), we get

$$
y(v-1)=0
$$

From this, if $v=1$, we arrive at the following system of equations:

$$
\begin{aligned}
& x+y+z=1 \\
& x+y+z=1 \\
& x+y+z=1
\end{aligned}
$$

which has infinitely many solutions, any triplet of numbers whose sum is 1. Therefore, we can assume that $v \neq 1$. If $y=0$ in (4), then the system of equations is:

$$
\begin{aligned}
x+z & =v \\
x+z & =v \\
x+v^{2} z & =v^{2}
\end{aligned}
$$

From the first equation, we get $x=v-z$, substituting this into the third equation:

$$
v-z+v^{2} z=v^{2}, \quad \text { that is } \quad z=\frac{v(v-1)}{v^{2}-1}
$$

if $v \neq-1$, then

$$
z=\frac{v(v-1)}{(v-1)(v+1)}
$$

Simplify the fraction by $(v-1)$, we get

$$
z=\frac{v}{v+1}, \quad \text { and } \quad x=v-\frac{v}{v+1}=\frac{v^{2}}{v+1}
$$

so the system of equations has a solution.

If, however, $v=-1$, then we arrive at the system of equations

$$
\begin{aligned}
& x+y+z=-1 \\
& x-y+z=-1 \\
& x+y+z=1
\end{aligned}
$$

The first and third equations are contradictory, so the system of equations has no solution when $v=-1$ and this is the only such value of $v$.