<s>[INST] Show that if $x+y+z=x^2+y^2+z^2=x^3+y^3+z^3=1$ then $xyz=0$ For any real numbers $x,y,z$ such that $x+y+z = x^2+y^2+z^2 = x^3+y^3+z^3 =1\\$ show that $x \cdot y \cdot z=0$.
I think that because $x \cdot y \cdot z = 0$ at least one of the three number should be equal to zero, but I'm stuck into relating the other things.
 [/INST] Consider $f(w)=(w-x)(w-y)(w-z).$ We have $f(w)=w^3-Aw^2+Bw-C$ where $A=x+y+z$ and $B=xy+yz+zx$ and $C=xyz.$ 
Suppose $x+y+z=x^2+y^2+z^2=x^3+y^3+z^3=1.$ We have $0=f(x)=f(y)=f(z).$ Hence $$0=f(x)+f(y)+f(z)=$$ $$=(x^3+y^3+z^3)-A(x^2+y^2+z^2)+B(x+y+z)-3C=$$ $$=1-A+B-3C=$$ $$=1-(x+y+z)+(xy+yz+zx)-3C=$$ $$=1-1+(xy+yz+zx)-3C=$$ $$=(xy+yz+zx)-3C.$$ Now $(xy+yz+zx)=\frac {1}{2} ((x+y+z)^2-(x^2+y^2+z^2))=\frac {1}{2} (1^2-1)=0 .$ 
Therefore $0=(xy+yz+zx)-3C=0-3C=0-3xyz.$
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