26.1. Real numbers $x$ and $y$ satisfy the equations $x^{3}-3 x^{2}+5 x=1, y^{3}-3 y^{2}+5 y=5$. Find $x+y$.

## 26.2. Periodic functions
26.1. Answer: $x+y=2$. The function $f(x)=x^{3}-3 x^{2}+5 x=(x-1)^{3}+$ $+2(x-1)+3$ is monotonically increasing, so for each real number $c$ the equation $f(x)=c$ has exactly one real solution.

The function $f(x)-3$ has the following property: if $x-1=$ $=1-y$, then $f(x)-3=-(f(y)-3)$. From the monotonicity of the function $f$ it follows that the converse is also true, i.e., if $f(x)-3=-(f(y)-3)$, then $x-1=1-y$, i.e., $x+y=2$. In the considered case, $f(x)=1$ and $f(y)=5$, so the specified equality holds.