Let x, x + 1, x + 2 be the three integers
Then, x(x + 1)(x + 2) = -1716
x(x2 + 3x + 2) = -1716
x3 + 3x2 + 2x + 1716 = 0
By trial and error, -13 is a root:
-13⌋ 1 3 2 1716
130 -1716
1 -10 132 0
x3 + 3x2 + 2x + 1716 = (x + 13)(x2 -10x + 132)
The roots of x2 - 10x + 132 are not real numbers
So, x = -13
The numbers are -13, -12, -11