2012 AMC 10A 第 17 题

Factor $a^3-b^3$

The equation becomes a quadratic in $\frac{a}{b}$

Recall that we can factor $$a^3 - b^3 = (a - b)(a^2 + ab + b^2).$$ Canceling out this factor gives us that $$\dfrac{a^2 + ab + b^2}{a^2 - 2ab + b^2} = \dfrac{73}{3}.$$

Cross-multiplying and rearranging gives us $$70a^2 - 149ab + 70b^2 = 0.$$ Since $b \neq 0,$ we can divide through by $b^2$ to get $$70\left(\dfrac{a}{b}\right)^2 - 149\dfrac{a}{b} + 70 = 0.$$

Applying the quadratic formula and noting that $a \gt b$ gives us that $\dfrac{a}{b} = \dfrac{10}{7}.$

Since $a$ and $b$ are relatively prime, we have that $a = 10$ and $b = 7.$ Their difference is $10 - 7 = 3.$

Thus, C is the correct answer.