Three consecutive numbers multiplied together give 3360, what are the numbers ?

Question

Of five consecutive multiples of three, the lowest is multiplied by two and four is added to give a number that is two thirds more than the average of the five numbers.
What is the highest of the consecutive numbers?

Answer 1

a*(a+1)*(a+2) = 3360
a*(a^2 + 3a + 2) = 3360
a^3 + 3a^2 + 2a - 3360 = 0 ----------- 1
factors of 3360 are 2^5, 3, 7 & 5
on closer examination of the factors by substituting on 1, a = 14 can be found to be the answer
So the numbers are: 14, 15 & 16.