The sum of the squares of three consecutive positive integers is 7805. what is the sum of the cubes of the three original integers?

398259

Step-by-step explanation:

Let from the given positive integers, x be the smallest integers,

Also, numbers are consecutive,

So, the second integer = x + 1,

Third integer = x + 2,

According to the question,

By middle term splitting,

By zero product property,

x-50 = 0 or x + 52 =0

⇒ x = 50 or x = -52 ( not possible )

Hence, numbers are 50, 51, 52,

∵ (50)³ + (51)³ + (52)³ = 125000 + 132651 + 140608 = 398259

77

step-by-step explanation: