Real numbers $a$ and $b$ satisfy the [equations](https://artofproblemsolving.com/wiki/index.php/Equation) $3^{a} = 81^{b + 2}$ and $125^{b} = 5^{a - 3}$. What is $ab$?
$\text{(A)}\ -60 \qquad \text{(B)}\ -17 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 60$
\[81^{b+2} = 3^{4(b+2)} = 3^a \Longrightarrow a = 4b+8\]
And
\[125^{b} = 5^{3b} = 5^{a-3} \Longrightarrow a - 3 = 3b\]
Substitution gives $4b+8 - 3 = 3b \Longrightarrow b = -5$, and solving for $a$ yields $-12$. Thus $ab = 60\ \mathrm{(E)}$.