2012 AMC 10A Problem 17

Factor the numerator: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$. Cancelling one factor of $a - b$,
$$\frac{a^2 + ab + b^2}{(a - b)^2} = \frac{73}{3}.$$
Let $d = a - b$. Since $a^2 + ab + b^2 = (a - b)^2 + 3ab = d^2 + 3ab$, the equation becomes
$$\frac{d^2 + 3ab}{d^2} = \frac{73}{3} \quad\Longrightarrow\quad 3d^2 + 9ab = 73d^2 \quad\Longrightarrow\quad 9ab = 70d^2.$$

Now use coprimality. Any common divisor of $d = a - b$ and $a$ also divides $b$, so $\gcd(d, a) = 1$; likewise $\gcd(d, b) = 1$. Hence $d$ shares no factor with $ab$, and $d^2 \mid 9ab$ forces $d^2 \mid 9$, so $d = 1$ or $d = 3$.

If $d = 1$, then $9ab = 70$, impossible. If $d = 3$, then $ab = 70$ with $a - b = 3$: the pair $a = 10$, $b = 7$ works, and $\gcd(10, 7) = 1$.

Check: $\frac{1000 - 343}{27} = \frac{657}{27} = \frac{73}{3}$.

The answer is $\boxed{\textbf{(C)}\ 3}$.

The difference-of-cubes identity plus the substitution $d = a - b$ turns a messy fraction into the clean relation $9ab = 70d^2$. The "relatively prime" hypothesis is there to be used: it makes $d$ coprime to $ab$, so all of $d^2$ must sit inside the $9$. Look for this pattern whenever a condition mentions $\gcd = 1$.

Test the answer choices in $9ab = 70d^2$: $d$ must be a multiple of $3$ (since $9 \mid 70d^2$), leaving only $d = 3$ among the choices, and $ab = 70$, $a - b = 3$ gives $(10, 7)$.

• Forgetting that gcd(a, b) = 1 implies gcd(a - b, ab) = 1, and so failing to conclude that d must divide 3 (or accepting d = 1 without checking that 9ab = 70 has no solution).
• Expanding $(a - b)^3$ fully and trying to solve a cubic instead of cancelling the common factor first.