Notice that you should use Vieta's relations for the x,y,z of a polynomial of third degree such that:

`at^3 + bt^2 + ct + d = 0`

`x+y+z = -b/a`

`xy+xz+yz = c/a`

`xyz = -d/a`

The problem provides the information that `x+y+z = 1, xy+xz+yz = -1` and `xyz = -1 ` such that:

`-b/a = 1 => -b = a => b = -a`

`c/a = -1 => c = -` a

`-d/a = -1 => d = a`

You should substitute -a or a for b,c,d such that:

`at^3 - at^2 - at + a = ` 0

Dividing by a yields:

`t^3 - t^2 - t +` 1 = 0

Sinc x,y,z are the roots of polynomial, hence, you should substitute x,y,z for t in equation above such that:

`x^3 - x^2 -x + 1 = 0`

`y^3 - y^2 - y + 1 = 0`

`z^3 - z^2 - z + 1 = 0`

Adding the equations yields:

`x^3 + y^3 + z^3 - (x^2 + y^2 + z^2) - (x + y + z) + 3 = 0`

You should evaluate `x^2 + y^2 + z^2` such that:

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

`x^2 + y^2 + z^2 = 1^2 - 2*(-1) = 3`

Substituting 3 for `x^2 + y^2 + z^2` and 1 for `x + y + z` , yields:

`x^3 + y^3 + z^3 - 3 - 1 + 3 = 0`

`x^3 + y^3 + z^3 = 1`

**Hence, evaluating the sum of cubes of roots x,y,z, under the given conditions, yields `x^3 + y^3 + z^3 = 1` .**

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