For a certain task, $B$ needs 6 more days than $A$, and $C$ needs 3 more days than $B$. If $A$ works for 3 days and $B$ works for 4 days, together they accomplish as much as $C$ does in 9 days. How many days would each take to complete the task individually?
If $A$ alone would complete the work in $x$ days, then $B$ alone would do it in $x+6$ days, and $C$ alone in $x+9$ days. Therefore, in one day, $A$ completes $\frac{1}{x}$ of the work, $B$ completes $\frac{1}{x+6}$ of the work, and $C$ completes $\frac{1}{x+9}$ of the work. According to the problem:

$$
\frac{3}{x}+\frac{4}{x+6}=\frac{9}{x+9}
$$

Eliminating the fractions:

$$
3(x+6)(x+9)+4 x(x+9)=9 x(x+6)
$$

And converting the products into sums and rearranging:

$$
2 x^{2}-27 x-162=0
$$

From which:

$$
x_{1}=18 \quad\left[x_{2}=-\frac{9}{2}\right]
$$

From the nature of the problem, only the positive root can be considered. Therefore, $A$ would complete the work alone in 18 days, $B$ in 24 days, and $C$ in 27 days.

Péter Eöllös (Tata, Eötvös József g. II. o. t.)