Check that 2 is a solution of the equation:

$$
(x+1)^{3}+(x+2)^{3}+(x+3)^{3}=(x+4)^{3}
$$

Does this equation have any other integer solutions?
$3^{3}+4^{3}+5^{3}=27+64+125=216=6^{3}$. Several techniques can be used to solve the equation: setting $x=2+k$, which, when expanded, gives a third-degree equation in $k$ of which one root is zero (this equation does not admit any other real root). But one can also expand directly: $(x+1)^{3}+(x+2)^{3}+(x+3)^{3}-(x+4)^{3}=$ $2 x^{3}+6 x^{2}-6 x-28$ : for this to be zero, it is necessary that $2 x$ divide the constant term -28, which limits the number of cases to study. One can also see that for $x \geq 3,2 x^{3}>28$ and $6 x^{2}>6 x$, so $(x+1)^{3}+(x+2)^{3}+(x+3)^{3}-(x+4)^{3}>0$ and for $x \leq-2,(x+1)^{3}+(x+2)^{3}<0$ and $(x+3)^{3}-(x+4)^{3}<0$, so $(x+1)^{3}+(x+2)^{3}+(x+3)^{3}-(x+4)^{3}<0$ : there remain four simple cases to study. Finally, one can note that modulo 3, as for any integer $n$, $n^{3} \equiv n$ $(\bmod 3),(x+1)^{3}+(x+2)^{3}+(x+3)^{3}-(x+4)^{3} \equiv(x+1)+(x+2)+(x+3)-(x+4)=2 x+2$, so $x$ must be congruent to 2 modulo 3.