find three consecutive integers whose product is 208 larger than the cube of the smallest integer

Question

the three consecutive integers are

Answer 1

n*(n+1)*(n+2)=n^3+208

n^3 +3n^2 +2n=n^3+208

3n^2 +2n=208

3n^2 +2n -208=0...yuze quadratik equashun

roots=8 or -8.66666666

yuze 8..8*9*10=720

8^3=512

720-512=208

Answer 2

The product of 16 and 14 less than a number is 208