# 2012 AMC 10A Problems/Problem 17

## Problem

Let $a$ and $b$ be relatively prime integers with $a>b>0$ and $\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}.$ 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 relatively prime, $a^3-b^3$ and $(a-b)^3$ are both integers as well. Then, for the given fraction to simplify to $\frac{73}{3}$, the denominator $(a-b)^3$ must be a multiple of $3.$ Thus, $a-b$ is a multiple of $3$. Looking at the answer choices, the only multiple of $3$ is $\boxed{\textbf{(C)}\ 3}$.

## 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 be seen upon squaring and summing the various factor pairs of $70$.

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

Remarks:

An alternate method of solving the system of equations involves solving the second equation for $a$, by plugging it into the first equation, and solving the resulting quartic equation with a substitution of $u = b^2$. The four solutions correspond to $(\pm10, \pm7), (\pm7, \pm10).$

Also, we can solve for $a-b$ directly instead of solving for $a$ and $b$: $a^2-2ab+b^2=149-2(70)=9 \implies a-b=3.$

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.

## Solution 3

The first step is the same as above which gives $\frac{a^2+ab+b^2}{a^2-2ab+b^2}=\frac{73}{3}$.

Then we can subtract $3ab$ and then add $3ab$ to get $\frac{a^2-2ab+b^2+3ab}{a^2-2ab+b^2}=\frac{73}{3}$, which gives $1+\frac{3ab}{(a-b)^2}=\frac{73}{3}$. $\frac{3ab}{(a-b)^2}=\frac{70}{3}$. Cross multiply $9ab=70(a-b)^2$. Since $a>b$, take the square root. $a-b=3\sqrt{\frac{ab}{70}}$. Since $a$ and $b$ are integers and relatively prime, $\sqrt{\frac{ab}{70}}$ is an integer. $ab$ is a multiple of $70$, so $a-b$ is a multiple of $3$. Therefore $a=10$ and $b=7$ is a solution. So $a-b=\boxed{\textbf{(C)}\ 3}$

## Solution 4

Slightly expanding, we have that $\frac{(a-b)(a^2+ab+b^2)}{(a-b)(a-b)(a-b)}=\frac{73}{3}$.

Canceling the $(a-b)$, cross multiplying, and simplifying, we obtain that

$0=70a^2-149ab+70b^2$. Dividing everything by $b^2$, we get that

$0=70(\frac{a}{b})^2-149(\frac{a}{b})+70$.

Applying the quadratic formula....and following the restriction that $a>b>0$....

$\frac{a}{b}=\frac{10}{7}$.

Hence, $7a=10b$.

Since they are relatively prime, $a=10$, $b=7$.

$10-7=\boxed{\textbf{(C)}\ 3}$.