431. Solve the equation:

$$
x^{3}+\sqrt{x}=y^{3}+\sqrt{y}
$$
$\triangle$ Given that $x$ and $y$ are non-negative. Let's consider two cases.

1) Suppose $y=x$. We get the set of solutions $(x ; x)$, where $x$ is any non-negative number.
2) Suppose $y \neq x$, for example, $y>x$.

The functions $z=x^{3}$ and $z=\sqrt{x}$ are increasing on the interval $[0 ;+\infty)$. Since the sum of two increasing functions is also an increasing function, it follows from the inequality $y>x$ that

$$
y^{3}+\sqrt{y}>x^{3}+\sqrt{x}
$$

which contradicts the condition. Therefore, in the second case, the given equation has no solutions.

Answer: $(x ; x)$, where $x \geq 0$.