Question

11. If x +y + z = 1, xy + yz + zx = - 1 and xyz = - 1, find the value x3 + y3 + z3

Answer 1

x3+y3+z3-3xyz=(x+y+z)(x2+y2+z2-xy-yz-zx)
x3+y3+z3+3=1(x2+y2+z2-(-1))
Now we have to find the value of x2+y2+z2
(x+y+z)^2=x2+y2+z2+2xy+2yz+2zx
1^2 =x2+y2+z2+2(xy+yz+zx)
1 =x2+y2+z2+2(-1)
1 =x2+y2+z2-2
1+2=x2+y2+z2
hence the value of x2+y2+z2=3
now put it's value in the equation
x3+y3+z3+3=1(3-(-1))
x3+y3+z3+3=1(3+1)
x3+y3+z3+3=4
x3+y3+z3=4-3
=1
Hence the value of x3+y3+z3=1