Question

if X + Y + Z is equals to 1, xy + yz +zx equals to -1 and xyz equals to - 1 then find x cube + y cube + Z cube

Answer 1

x^3+y^3+z^3=1 Ans.
x+y+z=1 (given)
xy+yz+zx= -1(")
xyz= -1(")

x+y+z=1
both side squaring in whole
(x+y+z)^2 = 1
x^2+y^2+z^2+2xy+2yz+2zx = (1)^2
x^2+y^2+z^2+2(-1) = 1
x^2+y^2+z^2 - 2=1
x^2+y^2+z^2 = 1+2
x^2+y^2+z^2 = 3

x^3+y^3+z^3+3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)
x^3+y^3+z^3+3(1)=1{3-1(-1)}
x^3+y^3+z^3+3=1(3+1)
x^3+y^3+z^3+3=1×4
x^3+y^3+z^3+3=4
x^3+y^3+z^3=4-3
x^3+y^3+z^3=1 Ans.

Answer 2

x^3+y^3+z^3
=6xyz+3x^2y+3x^z+3xy^2+3xz^2+3y^2z+3yz^2-(x+y+z)^3
=6(-1)+3xy(x+y)+3xz(x+z)+3yz(y+z)-(1)^3
=-6-1+3[xy(x+y)+xz(x+z)+yz(y+z)]
=-7+3[xy(x+y)+xz(x+z)+yz(y+z)]
=-7