Problema Solution
Find three consecutive integers whose product is 56 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 + 56
solving for 'x' we have a positive value of x=4
the integers are 4, 5 and 6