Question

sum of 3 numbers is 3 and the sum of its squares is 5 and the sum of its cubes is 9. find the numbers? ​

Answer 1

let three number is x, y and z
x+y+z=3
x^2+y^2+z^2=5
x^3+y^3+z^3=9
use
(x+y+z)^2=x^2+y^2+z^2+2 (xy+yz+zx)
9=5+2 (xy+yz+zx)
xy+yz+zx =2
now , x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)
=> 9-3xyz=3 (5-2)
xyz=0
hence one number must be zero
let z=0
then
x+y=3 (y=3-x)
x^2+(3-x)^2=5
2x^2-6x+4=0
x=1,2
so y=2,1
hence number is 1,2,0 or2,1,0