Question

if x+y+z=1,xy+yz+zx=-1,xyz=-1 then find value of x^3+y^3+z^3

Answer 1

$Identity:x ^{3} + {y}^{3} +z ^{3} -3xyz = (x+y+z)( x ^{2} +y ^{2} +z ^{2} -xy-yz-zx) \\ x+y+z=1,xy+yz+zx=-1,xyz=-1 \: \\ putting \: \: values \: we \: get \: \\ x ^{3} + {y}^{3} +z ^{3} -3( - 1) = (1)(x ^{2} + {y}^{2} + {z}^{2} + 1) \\ x ^{3} + {y}^{3} +z ^{3} + 3 \: = x ^{2} + {y}^{2} + {z}^{2} + 1 \: \\ x ^{3} + {y}^{3} +z ^{3} = x ^{2} + {y}^{2} + {z}^{2} + 1 - 3 \\ x ^{3} + {y}^{3} +z ^{3} = x ^{2} + {y}^{2} + {z}^{2} - 2 \: \: \: \: \: \: \: \: ....(1) \\ now \\ (x + y + z) ^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2(xy + yz + xz) \\ (1) ^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2( - 1) \\ 1 = {x}^{2} + {y}^{2} + {z}^{2} - 2 \\ {x}^{2} + {y}^{2} + {z}^{2} = 2 + 1 \\ {x}^{2} + {y}^{2} + {z}^{2} = 3 \\ put \: in \: equation \: \: (1) \\ x ^{3} + {y}^{3} +z ^{3} = 3 - 2 \\ x ^{3} + {y}^{3} +z ^{3} = \: 1$