3. Determine for which real number $p$ the equations

$$
\begin{array}{r}
x^{3}+x^{2}-36 x-p=0 \\
x^{3}-2 x^{2}-p x+2 p=0
\end{array}
$$

have a common root.

The school-based - closed part of the first round in category A takes place

## on Tuesday, December 3, 2002

so that it starts in the morning and the contestants have 4 hours of pure time to solve the problems. For each problem, a contestant can earn 6 points, and a successful solver is a student who scores 10 points or more. These details will be communicated to the students before the start of the competition.
Another solution. From the first equation, we can easily express $p=x^{3}+x^{2}-36 x$ and by substituting into the second equation, we get an equation (without the parameter $p$) that must be satisfied by the common root of both original equations:

$$
x^{3}-2 x^{2}-\left(x^{3}+x^{2}-36 x\right) x+2\left(x^{3}+x^{2}-36 x\right)=0 .
$$

After simplification, we get the equation $x^{4}-2 x^{3}+36 x^{2}-72 x=0$, whose roots can be easily determined (they are integers) by successive factorization:

$$
\begin{aligned}
x^{4}-2 x^{3}+36 x^{2}-72 x & =x\left[x^{2}(x-2)+36 x(x-2)\right]=x(x-2)\left(x^{2}-36\right)= \\
& =x(x-2)(x-6)(x+6) .
\end{aligned}
$$

We see that the common root must be one of the numbers $x_{1}=0, x_{2}=2, x_{3}=6, x_{4}=-6$. By substituting them into the original equations, we immediately find the corresponding values of $p$; they are the numbers 0, -60, and 36 (the last one corresponds to both roots $x_{3,4}= \pm 6$ ).