2018/004) Show that product of 4 consecutive natural numbers cannot be a perfect cube

Proof:
Here use use the fact that if p and q are co-primes and pq is a cube then both p and q both are cubes
there are 2 cases;
1) the lowest  number is odd.
the numbers are n, n+1, n+ 2, n+ 3 . and n+2 being odd lowest factor or $n+2 >=3$ hence it is co-prime to rest of the numbers
so n+2 is a cube and n(n+1)(n+3) is a cube
$n(n+1)(n+3) = n^3 + 4n^2 + 3n$
$n(n+1)(n+3) - (n+1)^3 = n^2 -1$
this is zero for n=1 which need to be checked and for n = 1 we have product 24 not a cube
for $n > 1$  $n(n+1)(n+3) > (n+1)^3$
$(n+2)^3 - (n^3 + 4n^2 + 3n)  = 2n^2 + 9n - 8 = 2n^2 + n + 8(n-1) > 0$
so it is between $(n+1)^3$ and$(n+2)^3$ and not a perfect cube
hence it is not a perfect cube for n odd

Now we see for n even

n+ 1 is odd and we need to show that $n(n+2)(n+3) = n^3 + 5n^2 + 6n$

so  $n(n+2)(n+3) - (n+1)^3  = 2n^2 + 3n-1 >0$

 $(n+2)^3 -  n(n+2)(n+3) = (n+2)((n+2)^2-n(n+3))$
$= (n+2)(n^2+4n+4 -n ^2- 3n) = (n+2)(n+4) >0$
so it is between $(n+1)^3$ and$(n+2)^3$ and not a perfect cube

  Hence it is not a perfect cube for any n