872. Solve the equation in natural numbers

$$
\sqrt[3]{x}+\sqrt[3]{y}=\sqrt[3]{1998}
$$
$\triangle$ Let's raise this equation to the third power:

$$
x+y+3 \sqrt[3]{x^{2} y}+3 \sqrt[3]{x y^{2}}=1998, \quad x+y+3 \sqrt[3]{x y}(\sqrt[3]{x}+\sqrt[3]{y})=1998
$$

What's next? Next, we replace the sum in the parentheses with $\sqrt[3]{1998}$:

$$
x+y+3 \sqrt[3]{1998 x y}=1998
$$

Factorize the number 1998 into prime factors: $1998=2 \cdot 3^{3} \cdot 37$. Therefore, for the cube root $\sqrt[3]{1998 x y}$ to be an integer, it is necessary that

$$
x y: 2^{2} \cdot 37^{2}
$$

Let $x$ be divisible by 2; then from the original equation, it is clear that $y$ is also divisible by 2. Similarly, if $x$ is divisible by 37, then $y$ is also divisible by 37. Therefore,

$$
x=74 a, \quad y=74 b
$$

where $a$ and $b$ are natural numbers. Substitute these expressions for $x$ and $y$ into the last equation:

$$
\begin{gathered}
74 a+74 b+3 \sqrt[3]{2 \cdot 3^{3} \cdot 37 \cdot 74 a \cdot 74 b}=2 \cdot 3^{3} \cdot 37 \\
74 a+74 b+3 \cdot 2 \cdot 3 \cdot 37^{3} \sqrt{a b}=2 \cdot 3^{3} \cdot 37, \quad a+b+9^{3} \sqrt{a b}=27
\end{gathered}
$$

From this, it follows that

$$
9 \sqrt[3]{a b}<27, \quad \sqrt[3]{a b}<3, \quad a b<27
$$

But since the product $a b$ must be a perfect cube, then $a b=1$ or $a b=8$

1) Let $a b=1$.

Then $a+b=18$. However, the system of equations

$$
\left\{\begin{array}{l}
a b=1 \\
a+b=18
\end{array}\right.
$$

has no solutions in natural numbers.

2) Let $a b=8$.

Then $a+b=9$. From the system of equations

$$
\left\{\begin{array}{l}
a b=8 \\
a+b=9
\end{array}\right.
$$

184
we find $a$ and $b: a=1, b=8$ or $a=8, b=1$. Therefore,

$$
x=74, y=74 \cdot 8=592 \text { or } x=592, y=74 \text {. }
$$

Answer: $(74 ; 592),(592 ; 74)$