For how many natural numbers $X(X+1)(X+2)(X+3)$ has exactly three different prime factors? 
For how many natural numbers $X(X+1)(X+2)(X+3)$ has exactly three different prime factors?

My attempt:
I have used a hit and trial approach. I found out that only for x=2 and x=3 this is happening. But how can I be sure of it? Is there any general approach?

you can use the following expission
$$\displaylines{
  x\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right) = x\left( {x + 3} \right)\left( {\left( {x + 1} \right)\left( {x + 2} \right)} \right) \cr 
   = x\left( {x + 3} \right)\left( {x^2  + 3x + 2} \right) \cr}$$
$$\left( {x\left( {x + 3} \right) + 1} \right)^2  - 1 = x\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right)$$
