Solve the following equation over the set of integer pairs:

$$
(x+2)^{4}-x^{4}=y^{3} \text {. }
$$
Solution. After performing the exponentiation and subtracting $x^{4}$:

$$
8 x^{3}+24 x^{2}+32 x+16=y^{3}
$$

Since 8 is a cube, let $8 b^{3}=y^{3}$, thus

$$
\begin{aligned}
8 x^{3}+24 x^{2}+32 x+16 & =8 b^{3} \\
x^{3}+3 x^{2}+4 x+2 & =b^{3}
\end{aligned}
$$

Transform the obtained equation:

$$
(x+1)^{3}+(x+1)=b^{3}
$$

Now let $a=x+1$, thus

$$
\begin{aligned}
a^{3}+a & =b^{3} \\
a\left(a^{2}+1\right) & =b^{3}
\end{aligned}
$$

Since $a$ and $a^{2}+1$ are relatively prime, $a\left(a^{2}+1\right)$ can only be a cube if both $a$ and $a^{2}+1$ are cubes. This is only possible if $a^{2}=0$ and $a^{2}+1=1$. Therefore, $x+1=0$, so $x=-1, y=0$.