Question

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

Answer 1

Answer:

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,

$x^2+(x+1)^2+(x+2)^2=7805$

$x^2+x^2+2x+1+x^2+4x+4=7805$

$3x^2+6x+5=7805$

$3x^2+6x+5-7805=0$

$3x^2+6x-7800=0$

$x^2+2x-2600$

By middle term splitting,

$x^2+(52-50)x-2600=0$

$x^2+52x-50x-2600=0$

$x(x+52)-50(x+52)=0$

$(x-50)(x+52)=0$

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