Problem source

Show that, if $n$ is an integer such that
$$(n-3)^3+n^3=(n+3)^3,\quad \quad {(*)}$$
then $n$ is even and $n^2$ is a factor of $54$. Deduce that there is no integer $n$ which satisfies the equation $(*)$.
Show that, if $n$ is an integer such that
$$(n-6)^3+n^3=(n+6)^3, \quad \quad{(**)}$$
then $n$ is even. Deduce that there is no integer $n$ which satisfies the equation $(**)$.

Solution source

\begin{align*}
&& n^3 &= (n+3)^3 - (n-3)^3 \\
&&&= n^3 + 9n^2+27n + 27 - (n^3 - 9n^2+27n-27) \\
&&&= 18n^2+54
\end{align*}

Therefore since $2 \mid 2(9n^2 + 27)$, $2 \mid n^3 \Rightarrow 2 \mid n$, so $n$ is even.

Since $n^2 \mid n^3$, $n^2 \mid 54 = 2 \cdot 3^3$, therefore $n = 1$ or $n = 3$. $(1-3)^3 + 1^3 < 0 < (1+3)^3$. So $n = 1$ doesn't work.

$(3 - 3)^3 + 3^3 < (3+3)^3$ so $n = 3$ doesn't work. Therefore there are no solutions.

\begin{align*}
&& n^3 &= (n+6)^3 - (n-6)^3 \\
&&&= n^3 + 18n^2 + 180n + 6^3 - (n^3 - 18n^2 + 180n - 6^3 ) \\
&&&= 36n^2+2 \cdot 6^3
\end{align*}

Therefore $n^2 \mid 2 \cdot 6^3 = 2^4 \cdot 3^3$, therefore $n = 1, 2, 3, 4, 6, 12$.

$n = 1$, $1^3 <36+2\cdot 6^3$
$n = 2$, $2^3 <36 \cdot 4 + 2 \cdot 6^3$ 
$n = 3$, $3^3 <36 \cdot 9 + 2 \cdot 6^3$ 
$n = 4$, $4^3 < 36 \cdot 16 + 2 \cdot 6^3$ 
$n = 6$, $6^3 < 36\cdot 6^2+ 2 \cdot 6^3$ 
$n = 12$, $12^3 < 36 \cdot 12^2 + 2 \cdot 6^3$ 

Therefore there are no solutions $n$ to the equation.

\textit{These are both special cases of Fermat's Last Theorem, when }$n = 3$