3. In the set of real numbers, solve the system

$$
\left\{\begin{array}{l}
x^{2}=y^{3} \\
x+y+\sqrt[5]{x y}=155
\end{array}\right.
$$
Solution. Clearly $y>0$. We introduce the substitution $y=z^{2}$, from which we get $x= \pm z^{3}$. For $x=z^{3}$, by substitution into the second equation, we obtain the equation $z^{3}+z^{2}+z=155$, which is subsequently equivalent to the equations:

$$
\begin{aligned}
& z^{3}-125+z^{2}-25+z-5=0 \\
& (z-5)\left(z^{2}+5 z+25\right)+(z-5)(z+5)+(z-5)=0 \\
& (z-5)\left(z^{2}+6 z+31\right)=0 \\
& (z-5)\left((z+3)^{2}+22\right)=0
\end{aligned}
$$

Now, from $(z+3)^{2}+22 \geq 22>0$ it follows that $z-5=0$, i.e., $z=5$. Finally, one solution to the system is $x=5^{3}=125, y=5^{2}=25$. For $x=-z^{3}$, by substitution into the second equation, we obtain the equation $-z^{3}+z^{2}-z=155$, which is subsequently equivalent to the equations:

$$
\begin{aligned}
& z^{3}-z^{2}+z+155=0 \\
& z^{3}+125-z^{2}+25+z+5=0 \\
& (z+5)\left(z^{2}-5 z+25\right)-(z-5)(z+5)+(z+5)=0 \\
& (z+5)\left(z^{2}-6 z+31\right)=0 \\
& (z+5)\left((z-3)^{2}+22\right)=0
\end{aligned}
$$

Now, from $(z-3)^{2}+22 \geq 22>0$ it follows that $z+5=0$, i.e., $z=-5$, which gives us $x=-(-5)^{3}=125, y=(-5)^{2}=25$.

Finally, the only real solution to the given system is $x=125, y=25$.