Let n – 1, n and n + 1 be positive consecutive integers such that (n – 1) ×n × (n + 1) = 16 × [(n – 1) + n + (n + 1)]. Simplifying and solving for n, we have n(n2 – 1) = 16 × 3n → n3 – n = 48n → n3 – 49n = 0

=> n(n2 – 49) = 0. So, n = 0 or n2 − 49 = 0 → (n + 7)(n – 7) = 0 => n = −7 or n = 7. Since n is a positive integer, it follows that the three consecutive numbers are 6, 7 and 8. The difference between their product and sum is 6 × 7 × 8 – (6 + 7 + 8) = 336 – 21 = 315.

 Let n – 1, n and n + 1 be positive consecutive integers such that (n – 1) ×n × (n + 1) = 16 × [(n – 1) + n + (n + 1)]. Simplifying and solving for n, we have n(n² – 1) = 16 × 3n → n³ – n = 48n → n³ – 49n = 0

=> n(n² – 49) = 0. So, n = 0 or n² − 49 = 0 → (n + 7)(n – 7) = 0 => n = −7 or n = 7. Since n is a positive integer, it follows that the three consecutive numbers are 6, 7 and 8. The difference between their product and sum is 6 × 7 × 8 – (6 + 7 + 8) = 336 – 21 = 315.