$$
x^{3}+a x^{2}+b x+6=0
$$

In the equation, determine $a$ and $b$ such that one root of the equation is 2 and another root is 3. Furthermore, calculate the third root.
If one root of the equation is 2 and the other root is 3, then

$$
8+4a+2b+6=0
$$

and

$$
27+9a+3b+6=0
$$

from which

$$
a=-4, b=1
$$

Since

$$
x_{1} x_{2} x_{3}=-6
$$

therefore

$$
x_{3}=-1
$$

(Klein Adolf, Székesfehérvár.)

Number of solutions: 37.