Question

设实数 $x$ 和 $y$ 满足 $x^4 y^5 + y^4 x^5 = 810$ 且 $x^3 y^6 + y^3 x^6 = 945$ 。求 $2x^3 + (xy)^3 + 2y^3$ 的值。

Let $x$ and $y$ be real numbers satisfying $x^4 y^5 + y^4 x^5 = 810$ and $x^3 y^6 + y^3 x^6 = 945.$ Evaluate $2x^3 + (xy)^3 + 2y^3.$

答案

答案：89

知识点：对称性（代数）换元法因式分解

难度评级：3160

解答：

两个方程可分解为 $x^4y^4(x + y) = 810$ 和 $x^3y^3(x^3 + y^3) = 945$ 。令 $s = x + y$ 、 $p = xy$ ，并使用 $x^3 + y^3 = s(s^2 - 3p)$ ，它们变为 $p^4 s = 810$ 和 $p^3 s\,(s^2 - 3p) = 945$ 。相除得 $\begin{aligned} &\frac{s^2 - 3p}{p} = \frac{945}{810} \\ &= \frac{7}{6}, \end{aligned}$ 所以 $6s^2 = 25p.$

把 $p = \frac{6s^2}{25}$ 代入 $p^4 s = 810$ ，得到 $\left(\frac{6}{25}\right)^4 s^9 = 810$ ，所以 $s^9 = 810 \cdot \frac{390625}{1296} = \frac{1953125}{8}$ ，这说明 $s^3 = \frac{125}{2}$ 。于是 $ps = \frac{6s^3}{25} = 15$ ，且 $p^3 = \frac{216 s^6}{25^3} = \frac{216 \cdot 15625/4}{15625} = 54$ 。

最后 $\begin{aligned} &2x^3 + (xy)^3 + 2y^3 \\ &= 2(s^3 - 3ps) + p^3 \\ &= 2\left(\frac{125}{2} - 45\right) + 54 \\ &= 35 + 54 = 89. \end{aligned}$

The equations factor as $x^4y^4(x + y) = 810$ and $x^3y^3(x^3 + y^3) = 945.$ With $s = x + y$ and $p = xy,$ using $x^3 + y^3 = s(s^2 - 3p),$ they become $p^4 s = 810$ and $p^3 s\,(s^2 - 3p) = 945.$ Dividing, $\begin{aligned} &\frac{s^2 - 3p}{p} = \frac{945}{810} \\ &= \frac{7}{6}, \end{aligned}$ so $6s^2 = 25p.$

Substituting $p = \frac{6s^2}{25}$ into $p^4 s = 810$ gives $\left(\frac{6}{25}\right)^4 s^9 = 810,$ so $s^9 = 810 \cdot \frac{390625}{1296} = \frac{1953125}{8},$ which means $s^3 = \frac{125}{2}.$ Then $ps = \frac{6s^3}{25} = 15$ and $p^3 = \frac{216 s^6}{25^3} = \frac{216 \cdot 15625/4}{15625} = 54.$

Finally $\begin{aligned} &2x^3 + (xy)^3 + 2y^3 \\ &= 2(s^3 - 3ps) + p^3 \\ &= 2\left(\frac{125}{2} - 45\right) + 54 \\ &= 35 + 54 = 89. \end{aligned}$

2015 AIME II 第 14 题

答案：89

难度评级：3160

解答：

其他年份的第 14 题