The roots of the equation $x^{3}-10 x+11=0$ are $u, v$, and $w$. Determine the value of

$$
\operatorname{arctg} u+\operatorname{arctg} v+\operatorname{arctg} w
$$
Solution. Let $\alpha=\operatorname{arctg} u, \beta=\operatorname{arctg} v$ and $\gamma=\operatorname{arctg} w$. Then the roots of the equation $x^{3}-10 x+11=0$ are $\operatorname{tg} \alpha$, $\operatorname{tg} \beta$ and $\operatorname{tg} \gamma$, and our task is to determine the sum $\alpha+\beta+\gamma$. The range of the function $x \rightarrow \operatorname{arctg} x$ is the interval $\left(-\frac{\pi}{2} ; \frac{\pi}{2}\right)$, so $-\frac{\pi}{2}<\alpha, \beta, \gamma<\frac{\pi}{2}$.

By repeatedly applying the addition theorem:

$$
\operatorname{tg}(\alpha+\beta+\gamma)=\frac{\operatorname{tg} \alpha+\operatorname{tg}(\beta+\gamma)}{1-\operatorname{tg} \alpha \cdot \operatorname{tg}(\beta+\gamma)}
$$

and from here, using the substitution $\operatorname{atg}(\beta+\gamma)=\frac{\operatorname{tg} \beta+\operatorname{tg} \gamma}{1-\operatorname{tg} \beta \cdot \operatorname{tg} \gamma}$, we get after rearrangement:

$$
\operatorname{tg}(\alpha+\beta+\gamma)=\frac{\operatorname{tg} \alpha+\operatorname{tg} \beta+\operatorname{tg} \gamma-\operatorname{tg} \alpha \cdot \operatorname{tg} \beta \cdot \operatorname{tg} \gamma}{1-(\operatorname{tg} \alpha \cdot \operatorname{tg} \beta+\operatorname{tg} \beta \cdot \operatorname{tg} \gamma+\operatorname{tg} \gamma \cdot \operatorname{tg} \alpha)}
$$

Expressing the right-hand side of (1) in terms of the roots of the given equation, we get:

$$
\operatorname{tg}(\alpha+\beta+\gamma)=\frac{u+v+w-u v w}{1-(u v+v w+w u)}
$$

The numerator and denominator of the fraction on the right-hand side of (1) are exactly the elementary symmetric polynomials of the roots of the equation $x^{3}-10 x+11=0$, which can be expressed in terms of the coefficients of the polynomial according to Viète's formulas:

$$
u+v+w=0, \quad u v+v w+w u=-10, \quad u v w=-11
$$

From this, $\operatorname{tg}(\alpha+\beta+\gamma)=\frac{11}{1+10}=1$, so $\alpha+\beta+\gamma=\frac{\pi}{4}+k \pi$, where $k$ is an integer whose value we still need to determine.

According to (2), the product of the roots is negative, so the number of negative roots is odd; the sum of the roots is zero, so it is not possible for all three roots to be negative: there is exactly one negative and two positive roots. Thus, one of $\alpha, \beta$ and $\gamma$ lies in the interval $\left(-\frac{\pi}{2} ; 0\right)$, and two lie in the interval $\left(0 ; \frac{\pi}{2}\right)$, so only $k=0$ is possible.

According to our results, $\operatorname{arctg} u+\operatorname{arctg} v+\operatorname{arctg} w=\frac{\pi}{4}$.

Remarks. 1. Many incomplete solutions were satisfied with the answer $\alpha+\beta+\gamma=\frac{\pi}{4}+k \pi$ without specifying the value of $k$.

2. Similarly, it can be proven that if $\operatorname{tg} \alpha, \operatorname{tg} \beta$ and $\operatorname{tg} \gamma$ are the roots of the polynomial $x^{3}+a_{1} x^{2}+a_{2} x+a_{3}$, then $\operatorname{tg}(\alpha+\beta+\gamma)=\frac{-a_{1}+a_{3}}{1-a_{2}}$, and by complete induction, it can be easily proven as a generalization of (1) that if the roots of the $n$-th degree polynomial $x^{n}+a_{1} x^{n-1}+a_{2} x^{n-2}+\ldots+a_{n}$ are $\operatorname{tg} \alpha_{1}, \operatorname{tg} \alpha_{2}, \ldots, \operatorname{tg} \alpha_{n}$, then

![](https://cdn.mathpix.com/cropped/2024_05_02_33238301cd05ee2284e0g-1.jpg?height=259&width=1037&top_left_y=1754&top_left_x=528)