1. In the set of natural numbers, solve the equation

$$
y^{3}=x^{3}+8 x^{2}-6 x+8
$$
Solution. From (1) it follows that for every real number $x$ we have

$$
y^{3}-(x+1)^{3}=y^{3}-x^{3}-3 x^{2}-3 x-1=5 x^{2}-9 x+7=5\left(x-\frac{9}{10}\right)^{2}+\frac{59}{20}>0
$$

i.e., $y>x+1$ and for every natural number $x$ we have

$$
y^{3}-(x+3)^{3}=y^{3}-x^{3}-9 x^{2}-27 x-27=-\left(x^{2}+33 x+19\right)<0
$$

i.e., $y<x+3$. We seek the solutions of equation (1) in the set of natural numbers, so from $x+1<y<x+3$ it follows that the only possibility is $y=x+2$. Substituting in (1), we obtain the equation

$$
(x+2)^{3}=x^{3}+8 x^{2}-6 x+8
$$

which is equivalent to the equation

$$
x^{2}-9 x=0
$$

Since $x \in \mathbb{N}$, from the last equation we get $x=9$, which means $y=11$.