3. It is known that the equations $x^{2}+a x+b=0$ and $x^{3}+b x+a=0$ have a common root and $a>b>0$. Find it.
Answer: -1.

Solution. Multiply the first equation by $x$ and subtract the second from it. The common root of the original equations will also be a root of the resulting equation

$$
\left(x^{3}+a x^{2}+b x\right)-\left(x^{3}+b x+a\right)=0 \quad \Longleftrightarrow \quad a\left(x^{2}-1\right)=0
$$

The last equation has two roots, which are 1 and -1. If the common root is $x=1$, then substituting it into each equation yields the equality $1+a+b=0$, which cannot hold given the condition $a>b>0$, leading to a contradiction.

If the common root is $x=-1$, then substituting it into the equations yields $1-a+b=0$, which does not contradict the condition.

Remark. Other methods of solution are possible.

Criteria. Only the answer - 0 points. Excess roots not eliminated - 2 points deducted. No verification of the root $x=-1$ for compliance with the condition - another 2 points deducted. One of the possible roots is missed - no more than 3 points. Complete solution - 7 points.