I followed a cube

$p$ is a prime number and $n$ , a natural one. ${ n }^{ 3 }-p=1\quad\Leftrightarrow\quad { n }^{ 3 }-1=p\quad \Leftrightarrow\quad p=\left( n-1 \right) \left( { n }^{ 2 }+n+1 \right)$

Because $p$ is prime and $n>0$ , $n-1>{ n }^{ 2 }+n+1$

So, $n>0$ , $n-1=1$

Finally, $n=2\quad\Leftrightarrow\quad { n }^{ 3 }=8\Leftrightarrow\ p={ n }^{ 3 }-1=7$