2017/015) Show that there are infinitely many n such that $6n+1$ and $6n-1$ are both are composites

we have for n > 2 $n^3-1= (n-1)(n^2+n+1)$ is composite and $n^3+1=(n+1)(n^2-n+1)$ is composite for n ,
so if we  choose $n= 36m^3$ we get $6n-1= (6m)^3-1$ and $6n+1= (6m)^3 + 1$ composites