If x+y+z=1, xy+yz+zx=-1, xyz=-1, then x^{3}+y | vedclass

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(D) We use the algebraic identity: $x^{3}+y^{3}+z^{3}-3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$.First,we find $x^{2}+y^{2}+z^{2}$ using the identity

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(D) We use the algebraic identity: $x^{3}+y^{3}+z^{3}-3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$.First,we find $x^{2}+y^{2}+z^{2}$ using the identity $(x+y+z)^{2} = x^{2}+y^{2}+z^{2}+2(xy+yz+zx)$.Substituting the given values: $(1)^{2} = x^{2}+y^{2}+z^{2} + 2(-1) \Rightarrow 1 = x^{2}+y^{2}+z^{2} - 2 \Rightarrow x^{2}+y^{2}+z^{2} = 3$.Now,substitute all values into the main identity:$x^{3}+y^{3}+z^{3} - 3(-1) = (1)(3 - (-1))$.$x^{3}+y^{3}+z^{3} + 3 = 1(3+1) = 4$.$x^{3}+y^{3}+z^{3} = 4 - 3 = 1$.

If $x+y+z=1, xy+yz+zx=-1, xyz=-1,$ then $x^{3}+y^{3}+z^{3}$ is

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