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:398259Step-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,[tex]x^2+(x+1)^2+(x+2)^2=7805[/tex][tex]x^2+x^2+2x+1+x^2+4x+4=7805[/tex][tex]3x^2+6x+5=7805[/tex][tex]3x^2+6x+5-7805=0[/tex][tex]3x^2+6x-7800=0[/tex][tex]x^2+2x-2600[/tex]By middle term splitting,[tex]x^2+(52-50)x-2600=0[/tex][tex]x^2+52x-50x-2600=0[/tex][tex]x(x+52)-50(x+52)=0[/tex][tex](x-50)(x+52)=0[/tex]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