Problema Solution
find three consecutive integers whose product is 208 larger than the cube of the smallest integer
Answer provided by our tutors
Let 'x' represent the smallest integer, then 'x+1' and 'x+2' represent the next two consecutive integers.
x(x+1)(x+2) = x^3 + 208
solving for x we have a positive value of x=8
The integers are 8, 9 and 10.