2012 AMC 10A Problems/Problem 17

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Problem

Let $a$ and $b$ be relatively prime integers with $a>b>0$ and $\frac{a^3-b^3}{(a-b)^3}$ = $\frac{73}{3}$. What is $a-b$?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution 1

Since $a$ and $b$ are both integers, so must $a^3-b^3$ and $(a-b)^3$. For this fraction to simplify to $\frac{73}{3}$, the denominator, or $a-b$, must be a multiple of 3. Looking at the answer choices, it is only possible when $a-b=\boxed{\textbf{(C)}\ 3}$.

Solution 2

Using difference of cubes in the numerator and cancelling out one $(a-b)$ in the numerator and denominator gives $\frac{a^2 + ab + b^2}{a^2 - 2ab + b^2} = \frac{73}{3}$.

Set $x = a^2 + b^2$, and $y = ab$. Then $\frac{x + y}{x - 2y} = \frac{73}{3}$. Cross multiplying gives $3x + 3y = 73x - 146y$, and simplifying gives $\frac{x}{y} = \frac{149}{70}$. Since $149$ and $70$ are relatively prime, we let $x = 149$ and $y = 70$, giving $a^2 + b^2 = 149$ and $ab = 70$. Since $a>b>0$, the only solution is $(a,b) = (10, 7)$, which can either be seen upon squaring and summing the various factor pairs of $70$.

Thus, the desired quantity $a - b = \boxed{\textbf{(C)}\ 3}$.

Note that if you double $x$ and double $y$, you will get different (but not relatively prime) values for $a$ and $b$ that satisfy the original equation.

See Also

2012 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AMC 10 Problems and Solutions
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