5. (3 points) The numbers $a, b$ and $c$ are distinct roots of the cubic polynomial $x^{3}+a x^{2}+b x-c$. Find them.

Answer: $a=1, b=-2, c=0$.
# Solution:

By applying Vieta's theorem, we obtain the system

$$
\left\{\begin{array}{l}
a+b+c=-a \\
a b+a c+b c=b \\
a b c=c
\end{array}\right.
$$

Let $c=0$. Then the third equation becomes an identity, and the first two equations transform into

$$
\left\{\begin{array}{l}
a+b=-a \\
a b=b
\end{array}\right.
$$

Since $b$ and $c$ are distinct, $b \neq 0$, and from the second equation, we get $a=1$. Therefore, $b=-2a=-2$.

If $c \neq 0$, from the third equation of the original system, we get $ab=1$, so $b=\frac{1}{a}$. Then from the first equation, $c=-2a-b=-2-\frac{1}{a}$.

The second equation of the original system becomes:

$$
\begin{gathered}
1+\left(a+\frac{1}{a}\right)\left(-2a-\frac{1}{a}\right)=-\frac{1}{a} \\
2a^2+2-\frac{1}{a}+\frac{1}{a^2}=0
\end{gathered}
$$

For negative $a$, all terms on the left side are positive.

For $a>1$, the inequalities $2+\frac{1}{a^2}>0$ and $2a^2-\frac{1}{a}>0$ hold.

For $0<a<1$, the inequalities $2a^2+2-\frac{1}{a}>0$ and $\frac{1}{a^2}-\frac{1}{a}>0$ hold.

Thus, the left side is always greater than zero, and the equation has no solutions.

Therefore, the only solution that works is the one from the case $c=0$.