21.9. (Czechoslovakia, 67). The numbers $a, b, c$ are three of the four roots of the polynomial $x^{4}-a x^{3}-b x+c$. Find all such triples of numbers $a, b, c$.
21.9. Suppose that $c=0$. Then the number $c$ is a root of the equation $x^{4}-a x^{3}-b x=0$, so it remains to find all polynomials of the form $x^{3}-a x^{2}-b$, for which the numbers $a$ and $b$ are roots. If we denote by $d$ the third root of such a polynomial, then by Vieta's theorem we will have $a+b+d=a$, i.e., $d=-b$, further, $a b + a d + b d = a b - b(a + b) = -b^{2} = 0$, i.e., $b=0$. Finally, we note that any triplet of numbers of the form $(a ; 0 ; 0)$ satisfies the condition of the problem. Now suppose that $c \neq 0$. Then not one of the four roots $a, b, c, d$ of the polynomial $x^{4}-a x^{3}-b x+c$ is equal to zero. By Vieta's theorem we have $a+b+c+d=a$, i.e., $d=-(b+c)$, further, $a b + a c + b c + a d + b d + c d =$

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

$a b c + a b d + a c d + b c d = a b c - (a b + a c + b c)(b + c) =$

$$
=-a\left(b^{2} + b c + c^{2}\right) - b^{2} c - b c^{2} = -b c(b + c) = b
$$

i.e.

$$
b^{2} = -c(b + c) = 1, \quad c^{2} + b c + 1 = 0
$$

finally,

$$
a b c d = -a b c(b + c) = a b = c,
$$

i.e., $a = c / b$. Therefore, only four sets of numbers $a, b, c$ are possible:

$$
\begin{gathered}
b_{1,2}=1, c_{1,2}=\frac{-1 \pm i \sqrt{3}}{2}, a_{1,2}=\frac{-1 \pm i \sqrt{3}}{2} \\
b_{8,4}=-1, c_{3,4}=\frac{1 \pm i \sqrt{3}}{2}, a_{3,4}=\frac{-1 \mp i \sqrt{3}}{2}
\end{gathered}
$$

each of which satisfies the condition of the problem. Thus, the triplet $(a ; b ; c)$ either has the form $(a ; 0 ; 0)$, where $a$ is any complex number, or coincides with one of the triplets:

$$
\begin{aligned}
& \left(\frac{-1+i \sqrt{3}}{2} ; 1 ; \frac{-1+i \sqrt{3}}{2}\right),\left(\frac{-1-i \sqrt{3}}{2} ; 1 ; \frac{-1-i \sqrt{3}}{2}\right) \\
& \left(\frac{-1-i \sqrt{3}}{2} ;-1 ; \frac{1+i \sqrt{3}}{2}\right),\left(\frac{-1+i \sqrt{3}}{2} ;-1 ; \frac{1-i \sqrt{3}}{2}\right)
\end{aligned}
$$