多项式实根

记

$$S=\sum_{u\in\{r,s,t\}}\frac{u-1}{u+1} =\sum\left(1-\frac{2}{u+1}\right)=3-2\sum\frac{1}{u+1}.$$

令 $y=u+1$，即考虑 $Q(y)=P(y-1)$ 的根为 $r+1,s+1,t+1$。

计算：

$$Q(y)=(y-1)^3-2007(y-1)+2002 = y^3-3y^2-2004y+4008.$$

设其根为 $y_1,y_2,y_3$，则（韦达）

$$\sum_{i<j}y_iy_j=-2004,\quad y_1y_2y_3=-4008.$$

因此

$$\sum\frac1{y_i}=\frac{\sum_{i<j}y_iy_j}{y_1y_2y_3}=\frac{-2004}{-4008}=\frac12.$$

故

$$S=3-2\cdot\frac12=\boxed{2}.$$

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英文解析

Recall

$$S=\sum_{u\in\{r,s,t\}}\frac{u-1}{u+1} =\sum\left(1-\frac{2}{u+1}\right)=3-2\sum\frac{1}{u+1}.$$

Let $y=u+1$, i.e., consider the roots of $Q(y)=P(y-1)$to be $r+1, s+1, t+1$.

Compute:

$$Q(y)=(y-1)^3-2007(y-1)+2002 = y^3-3y^2-2004y+4008.$$

Let its roots be $y_1, y_2, y_3$, then (by Vieta's formulas)

$$\sum_{i<j}y_iy_j=-2004,\quad y_1y_2y_3=-4008.$$

Therefore

$$\sum\frac1{y_i}=\frac{\sum_{i<j}y_iy_j}{y_1y_2y_3}=\frac{-2004}{-4008}=\frac12.$$

Thus

$$S=3-2\cdot\frac12=\boxed{2}.$$