## Task A-2.2.

Let the complex numbers $a, b$ and $c$ be the solutions of the equation $x^{3}-2 x+2=0$. Determine

$$
\frac{a+1}{a-1}+\frac{b+1}{b-1}+\frac{c+1}{c-1}
$$
## Solution.

We use Viète's formulas. Since $a, b$, and $c$ are the solutions of the equation $x^{3}-2 x+2=0$, we have

$$
a+b+c=0, \quad a b+b c+c a=-2, \quad a b c=-2
$$

It remains to express the given expression in terms of $a+b+c, a b+b c+c a$, and $a b c$:

$$
\begin{aligned}
\frac{a+1}{a-1}+\frac{b+1}{b-1}+\frac{c+1}{c-1} & =1+\frac{2}{a-1}+1+\frac{2}{b-1}+1+\frac{2}{c-1} \\
& =3+2 \cdot \frac{(b-1)(c-1)+(a-1)(c-1)+(a-1)(b-1)}{(a-1)(b-1)(c-1)} \\
& =3+2 \cdot \frac{(a b+b c+c a)-2(a+b+c)+3}{a b c-(a b+b c+c a)+(a+b+c)-1} \\
& =3+2 \cdot \frac{-2-0+3}{-2-(-2)+0-1}=1
\end{aligned}
$$