2. Let $a$ and $b$ be real numbers that satisfy the equations
$$
\frac{a}{b}+\frac{b}{a}=\frac{5}{2} \quad \text { and } \quad a-b=\frac{3}{2} \text {. }
$$

Find all possible values of $a^{2}+2 a b+b^{2}+2 a^{2} b+2 a b^{2}+a^{2} b^{2}$.
Answer: 0 and 81
Solution: From the 2nd equation, we have $a^{2}+b^{2}=\frac{9}{4}+2 a b$. Using this for the 1st equation, we have $\frac{a^{2}+b^{2}}{a b}=\frac{\frac{9}{4}+2 a b}{a b}=\frac{5}{2}$. and it can be solved that $a b=\frac{9}{2}$. Moreover, from the 2nd equation, we have $(a+b)^{2}=\frac{9}{4}+4 a b=\frac{81}{4}$. Thus, $a+b=\frac{9}{2}$ or $-\frac{9}{2}$ and $a^{2}+2 a b+b^{2}+2 a^{2} b+2 a b^{2}+a^{2} b^{2}=(a b+a+b)^{2}=81$ or 0 .