Question

X+y+z = 1; xy + yz +xz = -1 ; xyz = -1; x 3 + y 3 + z 3 =?

Answer 1

x+y+z=1------(1)
xy+yz+xz= -1--------(2)

xyz= -1------(3)

x3+y3+z3 = (x+y+z)[x2+y2+z2-(xy+yz+zx)]+3xyz
------(4)
do square of (1)

(x+y+z)2= x2+y2+z2+2(xy+yz+zx)

1^2= x2+y2+z2+2*(-1)

1+2 = x2+y2+z2

3=x2+y2+z2----(5)

substitute (1),(2),(3),and (5) in (4)

x3+y3+z3= 1[3-(-1)]+3*(-1)
=1*[3+1]-3
=4-3
=1

Answer 2

x3+y3+z3-3xyz=(x+y+z)(x2+y2+z2-xy-yz-zx)
x3+y3+z3=(x+y+z)(x2+y2+z2-xy-yz-zx)+3xyz
                =(-1)((x+y+z)2-2(xy+yz+zx))-1(xy+yz+zx))+3(xyz)
                =(-1)(1)(-2)(-1)(-1)(-1)+3(-1)
                =(-1+2+1-3)
                =(-1)
 Therefore, x3+y3+z3=(-1)