1. State the problem: If $a^3+b^3=217$ and $a+b=7$, find $ab$. 2. Use the identity for sum of cubes: $$a^3+b^3=(a+b)^3-3ab(a+b)$$ 3. Substitute the given values $a+b=7$ and $a^3+b^3=217$: $$217=7^3-3ab\cdot 7$$ 4. Simplify $7^3$: $$217=343-21ab$$ 5. Rearrange to isolate $ab$: $$21ab=343-217$$ $$21ab=126$$ 6. Divide both sides by $21$ (show canceling): $$\frac{21ab}{\cancel{21}}=\frac{126}{\cancel{21}}$$ $$ab=6$$ 7. Choose the correct option: $(b)\ 6$.