Task B-2.5. If $x_{1}$ and $x_{2}$ are the solutions of the equation $x^{2}+2013 x+1=0$, and $y_{1}$ and $y_{2}$ are the solutions of the equation $x^{2}+2014 x+1=0$, calculate the value of the expression $\left(x_{1}-y_{1}\right)\left(x_{2}-y_{2}\right)\left(x_{1}-y_{2}\right)\left(x_{2}-y_{1}\right)$.
## Solution.

According to Vieta's formulas, we have

$$
\begin{aligned}
& x_{1}+x_{2}=-2013, x_{1} \cdot x_{2}=1 \\
& y_{1}+y_{2}=-2014, y_{1} \cdot y_{2}=1
\end{aligned}
$$

In the given expression, we will first multiply the first and third, and the second and fourth parentheses, and then apply the previous equalities.

$$
\begin{aligned}
& \left(x_{1}-y_{1}\right)\left(x_{2}-y_{2}\right)\left(x_{1}-y_{2}\right)\left(x_{2}-y_{1}\right)= \\
& =\left(x_{1}^{2}-x_{1} y_{2}-x_{1} y_{1}+y_{1} y_{2}\right)\left(x_{2}^{2}-x_{2} y_{1}-x_{2} y_{2}+y_{1} y_{2}\right)= \\
& =\left(x_{1}^{2}-x_{1}\left(y_{1}+y_{2}\right)+y_{1} y_{2}\right)\left(x_{2}^{2}-x_{2}\left(y_{1}+y_{2}\right)+y_{1} y_{2}\right)= \\
& =\left(x_{1}^{2}+2014 x_{1}+1\right)\left(x_{2}^{2}+2014 x_{2}+1\right)= \\
& =(\underbrace{x_{1}^{2}+2013 x_{1}+1}_{0}+x_{1})(\underbrace{x_{2}^{2}+2013 x_{2}+1}_{0}+x_{2}) \\
& =x_{1} \cdot x_{2}=1
\end{aligned}
$$