Prove that product of four consecutive natural numbers can not be a perfect cube The question is 

Prove that product of four consecutive natural numbers can not be a perfect cube .

I really dont know what actually do to proceed with the question. However after seeing this related result for the product of four consecutive numbers Proving any product of four consecutive integers is one less than a perfect square
I tried with the same procedure of converting the product into a whole cube with a sum or difference of a constant . But i fail to do so. Any help or hint would be appreciated. Thank you
Edit: I just came to know the catalan's conjecture , so please if anyone could provide a proof for it or use some algebra to prove the question would be a great help.Also i think the question would not hold a very tough or lengthy proof since it is from a book recommended for students preparing for olympiads.
 A: First, note if a prime $p>3$ divides one of the numbers, it can't divide any other. So it must occur cubed, in the form $p^{3k}$ for some $k$.
Second, note that at least one of the four numbers must be co-prime to both $2$ and $3$; so this number contains only primes $p>3$, and it must therefore be a perfect cube.
And if one of the numbers is a perfect cube, then the product of the other three must also be a perfect cube. So now we have reduced the problem to showing that the product of three out of four consecutive numbers can't be a perfect cube.
There are three cases: $n(n+1)(n+2), n(n+1)(n+3),$ and $n(n+2)(n+3)$. We show by elementary algebra that each of these lies strictly between two consecutive cubes, except for $n(n+1)(n+3)=8$ when $n=1$; so none of them can be a perfect cube, except for this one case. But this case does not lead to a counter-example of the full statement, because the missing number $n+2=3$ is not a perfect cube.
Case 1: $n(n+1)(n+2)$
$n^3<n(n+1)(n+2)=n^3+3n^2+2n<n^3+3n^2+3n+1=(n+1)^3$
Case 2: $n(n+1)(n+3)$
$n(n+1)(n+3)-(n+1)^3=n^2-1$, so unless $n=1$, we have $(n+1)^3<n(n+1)(n+3)<(n+2)^3$.
Case 3: $n(n+2)(n+3)$
$(n+1)^3=n^3+3n^2+3n+1\underset{3n<2n^2+5}{<}n^3+5n^2+6=n(n+2)(n+3)\underset{n(n+3)<(n+2)^2}{<}(n+2)^3$
A: Assume for a contradiction that $n(n+1)(n+2)(n+3)$ is a cube, where $n$ is a positive integer; of course $n\gt1$ since $1\cdot2\cdot3\cdot4=24$ is not a cube.
Of the two middle factors $n+1$ and $n+2$, whichever one is odd is relatively prime to the other three factors, so it is a cube, and the product of the three remaining factors is a cube; i.e., either $n(n+1)(n+3)$ or $n(n+2)(n+3)$ must be a cube.
But you can easily verify that, since $n\gt1$,
$$(n+1)^3\lt n(n+1)(n+3)\lt n(n+2)(n+3)\lt(n+2)^3.$$
Since there are no cubes between $(n+1)^3$ and $(n+2)^3$, we have arrived at a contradiction.
A: Ok so let's see , I am taking a bit different approach here :
n(n+1)(n+2)(n+3)
For the above equation if we try to write the whole equation in the form of a single factor , (n+1) we can rewrite it as ,
(n+1)^4 + 2(n+1)^3 - (n+1)^2 - 2(n+1)
Now replacing (n+1) by t(where t is a natural number) we get ,
t^4 + 2 t^3 - t^2 - 2t
One taking factors common ,
(t^2 - 1)(t^2 + 2t)
=(t^2 - 1)[(t + 1)^2 - 1 ]
For this factor to be a cube root let's assume it to equal to p^3 (where p is a natural number).
Now by observation we can see that t^2 -1 < (t+1)^2 -1 , so the on comparing we can easily state that ,
(t^2 - 1) = p  and  (t+1)^2 - 1 = p^2 ( p^2 > p for any natural number)
Putting the original values again we get ,
(a) (n+1)^2 - 1 = p , which is not satisfied by any natural number values of n and p respectively .
(b) (n+2)^2 -1 = p^2 ,  which is again not satisfied by any natural number values of n and p respectively , which contradicts aur assumption.
Hence , it is proved that it cannot be a perfect cube.
