If #x+y+z=1#, #x^2+y^2+z^2=1# and #x^3+y^3+z^3=1#, prove that #xyz=0#?

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Answer 1 · 2018-03-04T22:14:02

Explanation in below.

#(x+y+z)^2=1^2#

#x^2+y^2+z^2+2*(xy+yz+xz)=1#

#1+2*(xy+yz+xz)=1#

#2*(xy+yz+xz)=0#, so #xy+yz+xz=0#

#(x+y+z)*(x^2+y^2+z^2)=1*1#

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

#1+x^2y+x^2z+y^2x+y^2z+z^2x+z^2y=1#

#xy*(x+y)+xz*(x+z)+yz*(y+z)=0#

#xy*(x+y+z-z)+xz*(x+y+z-y)+yz*(x+y+z-x)=0#

#xy*(1-z)+xz*(1-y)+yz*(1-x)=0#

#xy+xz+yz-3xyz=0#

Due to #xy+yz+xz=0#, #xyz# also must be #0#