Solve the equation in natural numbers: $x^{3}+y^{3}+1=3 x y$.

#
First method. Since $x>0$ and $y>0$, according to the Cauchy inequality, $\frac{x^{3}+y^{3}+1}{3} \geq \sqrt[3]{x^{3} y^{3} \cdot 1} \Leftrightarrow x^{3} + y^{3} + 1 \geq 3 x y$, and equality is achieved if and only if $x=y=1$.

Second method. If we move $3xy$ to the left side and replace 1 with $z$, the equation becomes $x^{3}+y^{3}+z^{3}-3xyz = 0$. According to problem $\underline{61105}$ g), the left side is equal to $(x+y+z)\left(x^{2}+y^{2}+z^{2}-xy-xz-yz\right)$.

Since $x, y$, and $z$ are natural numbers, only the second factor can be zero.

$x^{2}+y^{2}+z^{2}-xy-xz-yz=0 \Leftrightarrow 2x^{2}+2y^{2}+2z^{2}-2xy-2xz-2yz=0 \Leftrightarrow (x-y)^{2}+(y-z)^{2}+(z-x)^{2}=0$. Therefore, $x = y = z = 1$.

Third method. The equation can be written as

$(x+y)\left((x+y)^{2}-3xy\right)+1=3xy \Leftrightarrow \left((x+y)^{3}+1\right)-3xy(x+y)-3xy=0 \Leftrightarrow (x+y+1)\left((x+y)^{2}-(x+y)+1-3xy\right) = 0$.

Since $x$ and $y$ are natural numbers, $x+y+1>0$, so $(x+y)^{2}-(x+y)+1-3xy=0$.

Notice that $(x+y)^{2} \geq 4xy$, so $4xy-(x+y)+1-3xy \leq 0 \Leftrightarrow xy-x-y+1 \leq 0 \Leftrightarrow (x-1)(y-1) \leq 0$. For natural $x$ and $y$, the obtained inequality holds if and only if $x=1$ or $y=1$.

Substituting $y=1$ into the original equation, we get: $x^{3}-3x+2=0 \Leftrightarrow (x-1)\left(x^{2}+x-2\right)=0 \Leftrightarrow (x-1)^{2}(x+2)=0$. The last equation has only one natural root: $x=1$.

Since the original equation is symmetric with respect to $x$ and $y$, substituting $x=1$ into it, we get that $y = 1$.

Fourth method. Since the original equation is symmetric with respect to $x$ and $y$, it is sufficient to consider the case $x \geq y$.

Write the equation as $\left(x^{3}-3xy\right)+\left(y^{3}+1\right)=0$. Since $y$ is a natural number, $y^{3}+1>0$. Therefore, $x^{3}-3xy=x\left(x^{2}-3y\right)<0$. Thus,

$x^{2}-3y<0$ ( $x$ is natural). Considering that $x \geq y$, i.e., $-3x \leq -3y$, we get: $x^{2}-3x<0$. Therefore, $x < 3$.

This means that we only need to check three cases:

1) $x=y=2$ - not a solution;
2) $x=2, y=1$ - not a solution;
3) $x=y=1$ - is a solution.

## Answer

$(1,1)$

## Problem