Let's determine the numbers $a, b, c$ such that

$$
x^{3}-a x^{2}+b x-c=(x-a)(x-b)(x-c)
$$

is an identity.
I. solution. The desired identity holds if and only if, after multiplying out the right side and arranging by powers of $x$, the coefficients of the same powers of $x$ are equal on both sides:

$$
\begin{aligned}
-a & =-a-b-c \\
b & =a b+a c+b c=a(b+c)+b c \\
-c & =-a b c
\end{aligned}
$$

From (2), $b+c=0$, substituting this into (3) and reducing to 0,

$$
b(1-c)=0
$$

This can be satisfied in two ways:

I. if $b=0$, then from (2) $c=0$, and thus (4) is also satisfied, regardless of the coefficient $a$;

II. if $1-c=0$, i.e., $c=1$, then from (2) $b=-1$, and from (4) $a=-1$.

Tamas Hermann (Budapest, Fazekas M. Gymnasium, 3rd grade)

II. solution. Substitute $a$ for $x$ in (1), then we get

$$
a b-c=0, \quad c=a b
$$

Substituting this into (1) and replacing $x$ with $b$,

$$
b^{3}-a b^{2}+b^{2}-a b=b(b+1)(b-a)=0
$$

Finally, substitute $0$ for $x$, then using (5),

$$
-c=-a b c, c(a b-1)=a b(a b-1)=0
$$

(6) is satisfied in three cases:

I. if $b=0$, then from (5) $c=0$ and (1) holds for all $a$;

II. if $b=-1$, from (5) $c=-a$ and from (7) $a=0$ or $a=-1$. The former case does not yield an identity. $a=b=-1$, $c=1$ is, however, a solution to the problem;

III. if $b=a$, from (5) $c=a^{2}$ and from (7) $a^{2}\left(a^{2}-1\right)=0, a=0, a=1$ or $a=-1$.

In the first case, $a=b=c=0$, which is a special case of I. $a=b=c=1$ does not provide a solution again, while $a=b=-1, c=1$ is the solution found under II.