20. Let $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ be three different real solutions to the system of equations $\left\{\begin{array}{l}x^{3}-5 x y^{2}=21 \\ y^{3}-5 x^{2} y=28\end{array}\right.$. Find the value of $\left(11-\frac{x_{1}}{y_{1}}\right)\left(11-\frac{x_{2}}{y_{2}}\right)\left(11-\frac{x_{3}}{y_{3}}\right)$.
設 $\left(x_{1}, y_{1}\right)$ 、 $\left(x_{2}, y_{2}\right)$ 和 $\left(x_{3}, y_{3}\right)$ 分別為方程組 $\left\{\begin{array}{l}x^{3}-5 x y^{2}=21 \\ y^{3}-5 x^{2} y=28\end{array}\right.$ 三個不同的實
數解。求 $\left(11-\frac{x_{1}}{y_{1}}\right)\left(11-\frac{x_{2}}{y_{2}}\right)\left(11-\frac{x_{3}}{y_{3}}\right)$ 的值。
20. 1729
20. Dividing the first equation by the second, we get $\frac{x^{3}-5 x y^{2}}{y^{3}-5 x^{2} y}=\frac{3}{4}$. Dividing both the numerator and denominator on the left hand side by $y^{3}$, and writing $t$ for $\frac{x}{y}$, we have $\frac{t^{3}-5 t}{1-5 t^{2}}=\frac{3}{4}$, or
$$
4 t^{3}+15 t^{2}-20 t-3=0
$$

Clearly, every solution $\left(x_{0}, y_{0}\right)$ to the original system gives rise to a solution $t_{0}=\frac{x_{0}}{y_{0}}$ to $\left({ }^{*}\right)$. Moreover, two different solutions $\left(x_{0}, y_{0}\right)$ and $\left(x_{0}{ }^{\prime}, y_{0}{ }^{\prime}\right)$ to the original system give rise to two different solutions $t_{0}=\frac{x_{0}}{y_{0}}$ and $t_{0}{ }^{\prime}=\frac{x_{0}{ }^{\prime}}{y_{0}{ }^{\prime}}$ to $\left(^{*}\right.$ ). (This is because once $t_{0}$ is fixed, then so are $x_{0}$ and $y_{0}$; for instance the first equation together with $x_{0}=t_{0} y_{0}$ imply $y_{0}=\sqrt[3]{\frac{21}{t_{0}^{3}-5 t_{0}}}$.) Writing $t_{i}=\frac{x_{i}}{y_{i}}$, we need to find the value of $\left(11-t_{1}\right)\left(11-t_{2}\right)\left(11-t_{3}\right)$, where each $t_{i}$ is a different solution to $\left(^{*}\right)$, and hence $4 t^{3}+15 t^{2}-20 t-3=4\left(t-t_{1}\right)\left(t-t_{2}\right)\left(t-t_{3}\right)$. Consequently we have $\left(11-t_{1}\right)\left(11-t_{2}\right)\left(11-t_{3}\right)=\frac{4(11)^{3}+15(11)^{2}-20(11)-3}{4}=1729$.