Take a cube, add one, when is this a prime?

You can factorize $n^3+1$, so: $$ p= (n+1)(n^2-n+1)$$ so $n+1 = 1$ (so $n=0$) or $n^2-n+1=1$ (so $n(n-1)=0$...)

You can make your understanding stronger by the fact for any odd number $k > 1$ and $n > 1$, the number $p = n^k + 1$ is never a prime because it is divisible by $n+1$. The cubes or $k = 3$ is a special case of this.