A cubic equation has three real roots. The product of the roots is 2 more than their sum, the sum of their squares is 10, and the sum of their cubes is 6.

Which is this equation?
I. Solution. Let the equation be $x^{3}+a x^{2}+b x+c=0$, and the roots be $x_{1}, x_{2}, x_{3}$. According to Viète's formulas (the relationships between the roots and coefficients),

$$
\begin{aligned}
& a=-\left(x_{1}+x_{2}+x_{3}\right) \\
& b=x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3} \\
& c=-x_{1} x_{2} x_{3}
\end{aligned}
$$

It is easily verifiable that based on these,

$$
x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=a^{2}-2 b
$$

and

$$
x_{1}^{3}+x_{2}^{3}+x_{3}^{3}=-a^{3}+3 a b-3 c
$$

The information given in the problem, expressed in terms of $a, b, c$:

$$
\begin{aligned}
-c & =a+2 \\
a^{2}-2 b & =10 \\
-a^{3}+3 a b-3 c & =6
\end{aligned}
$$

From the first equation, $c=a-2$; from the second, $b=\frac{a^{2}}{2}-5$. Substituting these into the third equation:

$$
-a^{3}+3 a\left(\frac{a^{2}}{2}-5\right)-3(a-2)=6
$$

Rearranging:

$$
\begin{gathered}
-2 a^{3}+3 a\left(a^{2}-10\right)-6(a-2)=12 \\
a^{3}-36 a=0 \\
a(a-6)(a+6)=0
\end{gathered}
$$

This equation has three solutions: $a_{1}=0, a_{2}=6, a_{3}=-6$. The corresponding values of $b$ and $c$ based on the equations $b=\frac{a^{2}}{2}-5$ and $c=a-2$ are:

$$
\begin{array}{lll}
b_{1}=-5, & b_{2}=13, & b_{3}=13 \\
c_{1}=-2, & c_{2}=4, & c_{3}=-8
\end{array}
$$

For these three sets of numbers, the system of equations (1) is satisfied, but we need to check whether the corresponding equations indeed have three real roots. The three equations are:

$$
\begin{array}{r}
x^{3}-5 x-2=0 \\
x^{3}+6 x^{2}+13 x+4=0 \\
x^{3}+6 x^{2}+13 x-8=0
\end{array}
$$

It is known that the number of roots of the equation $y^{3}+p y+q=0$ can be determined from the sign of its discriminant, $\left(\frac{p}{3}\right)^{3}+\left(\frac{q}{2}\right)^{2}$:

If the discriminant is positive, there is one real root. If negative, there are three distinct real roots. If the discriminant is 0 and either $p$ or $q$ is non-zero, there is one double and one single real root. (If $p=q=0$, then $y=0$ is a triple root.)

The discriminant of equation (2) is $\left(\frac{-5}{3}\right)^{3}+\left(\frac{-2}{2}\right)^{2}$, which is negative, so it has three distinct real roots. For equations (3) and (4), we transform the left-hand side into a perfect cube to calculate the discriminant:

$$
\begin{aligned}
& (x+2)^{3}+(x+2)-6=0 \\
& (x-2)^{3}+(x-2)+2=0
\end{aligned}
$$

Their discriminants are $\left(\frac{1}{3}\right)^{3}+\left(\frac{-6}{2}\right)^{2}$ and $\left(\frac{1}{3}\right)^{3}+\left(\frac{2}{2}\right)^{2}$, respectively; both are positive. Therefore, these equations have only one root, and thus are not solutions to the problem.

The sought equation is therefore:

$$
x^{3}-5 x-2=0 \text {. }
$$

Remark. Some tried to prove that equations (3) and (4) have only one real root by showing that the derivative of the polynomial on the left-hand side is positive everywhere; this implies that the left-hand side is strictly monotonically increasing and cannot have two different roots.

This solution is incomplete in this form because it does not rule out the possibility of multiple roots. (It is not difficult to rule out multiple roots; for example, where there is a multiple root, the derivative is also 0, but in our case, it is positive.)

These solutions received 4 points.

II. Solution. Let the sought equation again be $x^{3}+a x^{2}+b x+c=0$, and the roots be $x_{1}, x_{2}, x_{3}$. According to the problem statement,

$$
\begin{aligned}
x_{1} x_{2} x_{3} & =x_{1}+x_{2}+x_{3}+2 ; \\
x_{1}^{2}+x_{2}^{2}+x_{3}^{2} & =10 ; \\
x_{1}^{3}+x_{2}^{3}+x_{3}^{3} & =6 .
\end{aligned}
$$

Subtract 3 times (5) from (7) and rearrange, using the identity

$$
u^{3}+v^{3}+w^{3}-3 u v w=(u+v+w)\left(u^{2}-u v+v^{2}-v w+w^{2}-w u\right)
$$

$$
\begin{aligned}
& x_{1}^{3}+x_{2}^{3}+x_{3}^{3}-3 x_{1} x_{2} x_{3}=-3\left(x_{1}+x_{2}+x_{3}\right) \\
& \left(x_{1}+x_{2}+x_{3}\right)\left(x_{1}^{2}-x_{1} x_{2}+x_{2}^{2}-x_{2} x_{3}+x_{3}^{2}-x_{3} x_{1}+3\right)=0
\end{aligned}
$$

The second factor is certainly not 0, because

$$
\begin{gathered}
x_{1}^{2}-x_{1} x_{2}+x_{2}^{2}-x_{2} x_{3}+x_{3}^{2}-x_{3} x_{1}+3= \\
=\frac{\left(x_{1}-x_{2}\right)^{2}}{2}+\frac{\left(x_{2}-x_{3}\right)^{2}}{2}+\frac{\left(x_{3}-x_{1}\right)^{2}}{2}+3 \geqslant 3
\end{gathered}
$$

Therefore, the equation can only be satisfied if $x_{1}+x_{2}+x_{3}=0$. In this case,

$$
\begin{aligned}
& a=-\left(x_{1}+x_{2}+x_{3}\right)=0 \\
& c=-x_{1} x_{2} x_{3}=-\left(x_{1}+x_{2}+x_{3}+2\right)=-2 \text { and } \\
& b=x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}=\frac{\left(x_{1}+x_{2}+x_{3}\right)^{2}-\left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\right)}{2}=\frac{0^{2}-10}{2}=-5 .
\end{aligned}
$$

The sought equation is therefore only $x^{3}-5 x-2=0$.

That it indeed has three roots can be proven as described in the first solution.

Equations (5) and (6) are satisfied by the choice of $b$ and $c$, and the third equation holds because the choice of $a$ makes equation (8) true, and (7) is obtained by adding 3 times (5) to (8).

The obtained equation is indeed a solution to the problem.