Problema Solution
Find three consecutive integers whose product is 16 larger than the cube of the smallest integer.
Answer provided by our tutors
Let the three consecutive integers be 'n', 'n+1', and 'n+2'. We can write:
(n)(n+1)(n+2)=(n^3)+16
Using Algebrator to solve for n, we get:
n=2
Therefore, the three integers are 2, 3, and 4.