3. Solve the system of equations

$$
\left\{\begin{array}{l}
x^{3}+2 x=y^{3}+2 y \\
x^{2014}+y^{2014}=2^{2015}
\end{array}\right.
$$
Solution. The function $f(t)=t^{3}+2 t$ is monotonically increasing on the entire number line as the sum of two monotonically increasing functions $f_{1}(t)=t^{3}$ and $f_{2}(t)=2 t$. Therefore, from the first equation we have $f(x)=f(y) \Leftrightarrow x=y$. The equality $X=Y$ can also be obtained by factoring the expression $x^{3}+2 x-y^{3}-2 y=0 \Rightarrow(x-y)\left(x^{2}+x y+y^{2}+2\right)=0$. Since $x^{2}+x y+y^{2}+2=(x+y / 2)^{2}+2+\frac{3}{4} y^{2}>0$, then $x=y$. Substituting $y=x$ into the second equation of the system, we get $x= \pm 2$. Answer. $x=y= \pm 2$

## Recommendations for checking

If only the answer is provided - 0 points.

If the solution is found but there is no proof of the non-existence of other solutions - 4 points.

Complete correct solution - 7 points.