Задача 54484 4 номер, пожалуйста ...
Условие
Решение
Все решения
[m]\begin{vmatrix}
\epsilon&\epsilon^2 &\epsilon^3 \\
\epsilon&0 &\epsilon \\
\epsilon^3&\epsilon^2 & \epsilon
\end{vmatrix}[/m]=(-1)^(1+2)[m]\begin{vmatrix}
\epsilon^2 &\epsilon^3 \\ \epsilon^2 & \epsilon
\end{vmatrix}[/m]+(-1)^(3+2)[m]\begin{vmatrix}
\epsilon&\epsilon^2 \\
\epsilon^3&\epsilon^2
\end{vmatrix}[/m]=[m]-(\epsilon^3-\epsilon^5)-(\epsilon^3-\epsilon^5)=2\epsilon^3\cdot (\epsilon^2-1)[/m]
[m]\epsilon=3-i[/m]
[m]\epsilon^2=(3-i)^2=9-6i+i^2=8-6i[/m]
[m]\epsilon^2-1=(3-i)^2=9-6i+i^2-1=7-6i[/m]
[m]\epsilon^3=(3-i)^3=3^3-3\cdot 3^2\cdot i+3\cdot3\cdot i^2-i^3=27-27i-9+i=18-26i[/m]
[m]2\epsilon^3\cdot (\epsilon^2-1)=2\cdot (18-26i)\cdot (7-6i)=[/m] упрощайте...