Given that $a+b=3$ and $a^3+b^3=81$, find $ab$.
I notice that the expressions $a+b$ and $a^3+b^3$ are related by a cubic identity: $(a+b)^3=a^3+b^3+3ab(a+b)$. If I plug in the given values, I get $3^3=81+3ab(3)$, which simplifies to $27=81+9ab$. Subtracting $81$ from both sides, I get $-54=9ab$. Dividing by $9$, I get $-6=ab$. Therefore, $ab=-6$. 