## Task B-4.1.

In the set of complex numbers, solve the equation

$$
(x+5)^{5}=x^{5}+5^{5}
$$
## Solution.

By applying the binomial theorem, the given equation can be written in the form

$$
x^{5}+5 \cdot x^{4} \cdot 5+10 \cdot x^{3} \cdot 5^{2}+10 \cdot x^{2} \cdot 5^{3}+5 \cdot x \cdot 5^{4}+5^{5}=x^{5}+5^{5}
$$

from which, after simplification, we obtain the equation $25 x^{4}+250 x^{3}+1250 x^{2}+3125 x=0$.

By factoring the expression on the left side of the equation, we obtain the following sequence of equalities:

$$
\begin{aligned}
25 x\left(x^{3}+10 x^{2}+50 x+125\right) & =0 \\
25 x\left((x+5)\left(x^{2}-5 x+25\right)+10 x(x+5)\right) & =0 \\
25 x(x+5)\left(x^{2}+5 x+25\right) & =0
\end{aligned}
$$

From the last equation, we find that the solutions to the given equation are:

$$
x_{1}=0, x_{2}=-5, x_{3}=\frac{-5+5 \sqrt{3} i}{2}, x_{4}=\frac{-5-5 \sqrt{3} i}{2}
$$