AMC10 2012 A

AMC10 2012 A · Q17

AMC10 2012 A · Q17. It mainly tests Factoring, GCD & LCM.

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?

设$a$和$b$是互质的正整数，且$a > b > 0$，$\frac{a^3 - b^3}{(a - b)^3} = \frac{73}{3}$。求$a - b$的值。

(A) 1 1

(B) 2 2

(D) 4 4

(E) 5 5

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Note that $\dfrac{a^3-b^3}{(a-b)^3}=\dfrac{a^2+ab+b^2}{a^2-2ab+b^2}.$ Hence the given equation may be written as $3a^2+3ab+3b^2=73a^2-146ab+73b^2$. Combining like terms and factoring gives $(10a-7b)(7a-10b)=0$. Because $a>b$, and $a$ and $b$ are relatively prime, $a=10$ and $b=7$. Thus $a-b=3$.

答案（C）：注意 $\dfrac{a^3-b^3}{(a-b)^3}=\dfrac{a^2+ab+b^2}{a^2-2ab+b^2}.$ 因此，所给方程可写为 $3a^2+3ab+3b^2=73a^2-146ab+73b^2$。合并同类项并因式分解得 $(10a-7b)(7a-10b)=0$。由于 $a>b$，且 $a$ 与 $b$ 互素，所以 $a=10$，$b=7$。因此 $a-b=3$。

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