2012 AMC 10A Problem 17

1. The top factors: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$. Cancel one factor of $a - b$ from the top and bottom:
$$\frac{a^2 + ab + b^2}{(a - b)^2} = \frac{73}{3}.$$
2. The bottom is about $a - b$, so write the top in terms of $a - b$ too. Since $(a - b)^2 = a^2 - 2ab + b^2$, we get $a^2 + ab + b^2 = (a - b)^2 + 3ab$.
3. Call $d = a - b$. The equation becomes
$$\frac{d^2 + 3ab}{d^2} = \frac{73}{3}.$$
Cross-multiply: $3d^2 + 9ab = 73d^2$, so
$$9ab = 70d^2.$$
4. Now use the fact that $a$ and $b$ have no common factor. If a prime divided both $d = a - b$ and $a$, it would also divide $b = a - d$. That cannot happen. So $d$ shares no prime factor with $a$, and in the same way none with $b$.
5. The right side $70d^2$ is a multiple of $d^2$, so $9ab$ must be too. Since $d$ shares nothing with $a$ or $b$, all of $d^2$ has to fit inside the $9$. So $d^2$ divides $9$, which means $d = 1$ or $d = 3$.
6. If $d = 1$: $9ab = 70$, but $70$ is not a multiple of $9$. Impossible.
7. If $d = 3$: $9ab = 630$, so $ab = 70$, with $a - b = 3$. Try the ways to write $70$ as a product: $10 \times 7$ works, since $10 - 7 = 3$. And $10$ and $7$ share no factor.
8. Check: $\dfrac{10^3 - 7^3}{3^3} = \dfrac{1000 - 343}{27} = \dfrac{657}{27} = \dfrac{73}{3}$.

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

The difference-of-cubes factoring, together with naming $d = a - b$, turns a messy fraction into the clean equation $9ab = 70d^2$. The condition that $a$ and $b$ share no factor is in the problem to be used: it means $d$ shares no factor with $ab$, so $d^2$ must divide the $9$. When a problem says two numbers are relatively prime, ask what that forces about their sum or difference.

Test the choices in $9ab = 70d^2$. The left side is a multiple of $9$, and $70$ has no factor of $3$, so $d^2$ must be a multiple of $9$. Among the choices, only $d = 3$ works. Then $ab = 70$ and $a - b = 3$, which gives $a = 10$, $b = 7$.

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 $\boxed{\textbf{(C)}\ 3}$.