2012 AMC 10A Problems/Problem 17

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

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

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.

2012 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

Page

Toolbox

Search

2012 AMC 10A Problems/Problem 17

Contents

Problem

Solution 1

Solution 2

See Also