3. In the set of complex numbers, solve the equation

$$
(x-1)(x-2)(x-4)(x-5)=40
$$
Solution. Since

$$
(x-1)(x-5)=x^{2}-6 x+5,(x-2)(x-4)=x^{2}-6 x+8
$$

the equation takes the form

$$
\left(x^{2}-6 x+5\right)\left(x^{2}-6 x+8\right)=40
$$

Setting $x^{2}-6 x+5=y$, we get the equation $y(y+3)=40$ whose solutions are $y_{1}=5$ and $y_{2}=-8$. Solving now the equations

$$
x^{2}-6 x+5=5, \quad x^{2}-6 x+5=-8
$$

we obtain the roots of the given equation: $x_{1}=0, x_{2}=6, x_{3}=3+2 i, x_{4}=3-2 i$.