3B. Let $a$ and $b$ be real numbers such that $\frac{a}{b}+\frac{b}{a}=\frac{5}{2}$ and $a-b=\frac{3}{2}$. Determine all possible values of the expression

$$
A=a^{2}+2 a b+b^{2}+2 a^{2} b+2 a b^{2}+a^{2} b^{2}
$$
Solution. From $a-b=\frac{3}{2}$ it follows that $a^{2}-2 a b+b^{2}=\frac{9}{4}$, i.e., $a^{2}+b^{2}=\frac{9}{4}+2 a b$. Substituting into $\frac{a^{2}+b^{2}}{a b}=\frac{5}{2}$, we get $a b=\frac{9}{2}$. Further, if we add $4 a b$ to both sides of the equation $a^{2}-2 a b+b^{2}=\frac{9}{4}$, we obtain the equivalent equation $(a+b)^{2}=\frac{9}{4}+4 a b$. Substituting $a b=\frac{9}{2}$, we get $(a+b)^{2}=\frac{81}{4}$, from which we find $a+b=\frac{9}{2}$ or $a+b=-\frac{9}{2}$.

We have

$$
\begin{aligned}
A & =a^{2}+2 a b+b^{2}+2 a^{2} b+2 a b^{2}+a^{2} b^{2} \\
& =(a+b)^{2}+2 a b(a+b)+(a b)^{2} \\
& =(a+b+a b)^{2}
\end{aligned}
$$

so for $a b=\frac{9}{2}$ and $a+b=\frac{9}{2}$, we get $A=\left(\frac{9}{2}+\frac{9}{2}\right)^{2}=81$, and for $a b=\frac{9}{2}$ and $a+b=-\frac{9}{2}$, we get $A=\left(\frac{9}{2}-\frac{9}{2}\right)^{2}=0$. Finally, under the given conditions, the expression A takes only two values, namely 0 and 81.

4AB. Given is a parallelogram $A B C D$. Let $E$ be a point on the ray $D B$ such that $A B$ is the bisector of $\measuredangle C A E$ and let $F=C E \cap A B$. Prove that $\frac{\overline{A B}}{\overline{B F}}-\frac{\overline{A C}}{\overline{A E}}=1$.