It is not possible in either case. Proof:
(n+1)(n+2) = n² + 3n + 2 and n(n+3) = n² + 3n. So their product is (n2 + 3n + 1)² - 1. Hence n(n + 1)(n + 2)(n + 3) is 1 less than a square, so it cannot be a square.
One of n+1, n+2 must be odd. Suppose it is n+1. Then n+1 has no factor in common with n(n + 2)(n + 3), so n(n + 2)(n + 3) = n³ + 5n² + 6n must be a cube. But (n + 1)³ = n³ + 3n² + 3n + 1 < n³ + 5n² + 6n < n³ + 6n² + 12n + 8 = (n + 2)³, so n³ + 5n² + 6n cannot be a cube.
Similarly, suppose n + 2 is odd. Then it has no factor in common with n(n + 1)(n + 3) = n³ + 4n² + 3n, so n³ + 4n² + 3n must be a cube. But for n >= 2, (n + 1)³ < n³ + 4n² + 3n < (n + 2)³, so n³ + 4n³ + 3n cannot be a cube for n >= 2. The case n = 1 is checked by inspection: 24 is not a cube. |