If $a + b + c = 6$ and $a^2 + b^2 + c^2 = 14$, find $ab + bc + ca$.

**Identity.** $(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$. **Plug in.** $36 = 14 + 2(ab+bc+ca) \;\Rightarrow\; ab+bc+ca = \dfrac{22}{2} = 11$. **Why option B is tempting.** Subtracting $14$ from $36$ gives $22$, and half-remembered identities lead some to write $22$ or $10$. The factor of two on the cross-product term is the load-bearing detail.