Find three consecutive positive integer which …

Let three consecutive positive integers be x, x + 1 and x + 2
As per given condition  product of three consecutive positive integers is equal to sixteen times their sum.
{tex}\therefore{/tex}  (x)(x +1)(x + 2) = 16(x + x + 1 + x + 2)
{tex}\Rightarrow{/tex}(x² + x)(x + 2) = 16(3x + 3)
{tex}\Rightarrow{/tex}x³ + 2x² + x² + 2x = 48x + 48

{tex}\Rightarrow{/tex} x³ + 3x² - 46x - 48 = 0
When x = -1, we have
LHS  = (-1)³ + 3(- l)² - 46  (-1) -48
= -1 + 3 + 46-48 = 0
{tex}\therefore{/tex}(x + 1) is a factor
For other factors 

{tex}\therefore{/tex}x³ + 3x² - 46x - 48 = 0

{tex}\Rightarrow{/tex}(x + 1)(x² + 2x - 48) = 0
{tex}\Rightarrow{/tex} (x + 1) (x + 8)(x - 6) = 0

{tex}\Rightarrow{/tex} x = -1, x = -8, x = 6.
Rejecting -ve values,therefore, x = 6

{tex}\therefore{/tex}Positive integers are 6, 7 and 8.