Two reals $a$ and $b$ are such that $a+b=7$ and $a^3+b^3=91$. Compute $ab$.
To solve for $ab$, I need to somehow relate it to $a+b$ and $a^3+b^3$. I recall that $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$. Maybe I can use this identity to eliminate one of the variables. If I plug in $a+b=7$ and $a^3+b^3=91$, I get $7^3 = 91 + 3ab(7)$. Simplifying, I get $343 - 91 = 21ab$. Dividing by 21, I get $ab = \frac{252}{21} = 12$. 