14. [40] Find the prime factorization of
$$
2006^{2} \cdot 2262-669^{2} \cdot 3599+1593^{2} \cdot 1337 \text {. }
$$
(No proof is necessary.)
Answer: $2 \cdot 3 \cdot 7 \cdot 13 \cdot 29 \cdot 59 \cdot 61 \cdot 191$
Solution: Upon observing that $2262=669+1593,3599=1593+2006$, and $1337=$ $2006-669$, we are inspired to write $a=2006, b=669, c=-1593$. The expression in question then rewrites as $a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)$. But, by experimenting in the general case (e.g. setting $a=b$ ), we find that this polynomial is zero when two of $a, b, c$ are equal. Immediately we see that it factors as $(b-a)(c-b)(a-c)$, so the original expression is a way of writing $(-1337) \cdot(-2262) \cdot(3599)$. Now, $1337=7 \cdot 191$, $2262=2 \cdot 3 \cdot 13 \cdot 29$, and $3599=60^{2}-1^{2}=59 \cdot 61$.