## Task A-4.2.

Let $x_{1}, x_{2}$, and $x_{3}$ be distinct roots of the polynomial $P(x)=x^{3}-3 x+1$. Determine

$$
\frac{1}{x_{1}^{2}-3}+\frac{1}{x_{2}^{2}-3}+\frac{1}{x_{3}^{2}-3}
$$
## First Solution.

For the root $x_{1}$ of the polynomial $P$, we have

$$
\begin{aligned}
x_{1}^{3}-3 x_{1}+1 & =0 \\
x_{1}^{3}-3 x_{1} & =-1 \\
x_{1}^{2}-3 & =-\frac{1}{x_{1}} \\
\frac{1}{x_{1}^{2}-3} & =-x_{1} .
\end{aligned}
$$

Similarly, this holds for $x_{2}$ and $x_{3}$, so for the desired expression we have

$$
\frac{1}{x_{1}^{2}-3}+\frac{1}{x_{2}^{2}-3}+\frac{1}{x_{3}^{2}-3}=-\left(x_{1}+x_{2}+x_{3}\right)
$$

By Vieta's formulas, $x_{1}+x_{2}+x_{3}=0$, so the expression is equal to zero.