For a fixed positive integer n, if
$D = \left|\begin{array}{ccc} n! & (n + 1)! & (n + 2)! \\ (n + 1)! & (n + 2)! & (n + 3)! \\ (n + 2)! & (n + 3)! & (n + 4)! \end{array} \right|$
show that $\left(\dfrac{D}{(n!)^{3}} - 4 \right)$ is divisible by $n$.
Any ideas on how to go about solving this??
Thank You in advance.