2017/020) Find n such that both $n+3$ and $n^2+3$ are perfect cubes

If $(n+3)$  and $n^2+3$ are perfect cubes then their product is a perfect cube say $x^3$

$x^3 = (n+3)(n^2+3) = n^3 + 3n^3 + 3n + 9 = (n+1)^3 + 8 = (n+1)^3 + 2^3$

$x^3 = (n+1)^3 + 2^3$ this does not have a non trivial solution

so x = 0 or n = -1 and n= -1 => n+3 is 2 which is not a cube
x = 0 => n+1 = - 2 so $n^2  + 3 =  12$ which is not a cube

so no solution