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.
 A: a nice problem! squaring the first equation we obtain
$$xy+xz+yz=0$$
raise to the power $3$ the last equation we have
$$x^2(y+z)+y^2(x+z)+z^2(x+y)+xyz=0$$
and this can be written as
$$x^2(1-x)+y^2(1-y)+z^2(1-z)+xyz=0$$
and this is
$$x^2+y^2+z^2-x^3-y^3-z^3+xyz=0$$ therefore $xyz=0$
A: Let $q(t)=(t-x)(t-y)(t-z)$ and $p_n=x^n+y^n+z^n$. Power sums give a base of the ring of symmetric polynomials in the variables $x,y,z$, and by Newton's identities
$$ e_1 = x+y+z = p_1=1, $$
$$ 2e_2 = 2xy+2yz+2xz = p_1^2-p_2 = 0, $$
$$3e_3 = 3xyz = e_2 p_1 - e_1 p_2 + p_3 = 0, $$
so $q(t) = t^3-t^2$ and $x,y,z$ are given by a permutation of $0,0,1$.
More generally, if we have $n$ variables $x_1,\ldots,x_n$ and the power sums from $p_1$ to $p_n$ all equal $1$, then $x_1,\ldots,x_n$ are given by a permutation of $0,0,\ldots,1$, since $t^n-t^{n-1}$ is the only monic polynomial associated to such power sums.
A: $$x^3+y^3+z^3-3xyz=(x+y+z)\{(x+y+z)^2-3(xy+yz+zx)\}$$
Now $$2(xy+yz+zx)=(x+y+z)^2-(x^2+y^2+z^2)=?$$
A: For $x:0\neq|x|<1$ we have $x^3<x^2$. Therefore if we have 3 numbers $x_1^2+x_2^2+x_3^2=1$ and $x_1x_2x_3\neq0$ we have that $x_i\neq0$ $\forall i$. Therefore for every $i$ we have the strict inequallity $x_i^3<x_i^2$ therefore $x_1^3+x_2^3+x_3^3<1$. Therefore one of the $x_i$ is $0$ .
A: $$2(xy+xz+yz)=(x+y+z)^2-(x^2+y^2+z^2)=0,$$ which says $xy+xz+yz=0$.
Thus, $$1=(x+y+z)^3=x^3+y^3+z^3+3(x+y+z)(xy+xz+yz)-3xyz=1-3xyz$$ and we are done!
A: 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 $0=(xy+yz+zx)-3C=0-3C=0-3xyz.$
A: We have $$(x+y+z)(x^2-xy+y^2+z^2)=x^3+y^3+z^3+xz^2+yz^2+x^2z-xyz+y^2z$$ so $$x^2-xy+y^2+z^2=1+xz^2+yz^2+x^2z-xyz+y^2z$$ so $$-xy=xz(x+z)+yz(y+z)-xyz$$ giving $$xyz=xy+xz-xyz+yz-xyz\implies 3xyz=xy+xz+yz$$ Now $$1=(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)\implies xy=xz+yz=0$$ Hence $$\boxed{3xyz=0\implies xyz=0.}$$
A: \begin{align}
\left( \sum_{cyc} x \right) \left( \sum_{\text{cyc}} x^2
\right) &= \left( \sum_{\text{cyc}} x^3 \right) = 1 \\
\sum_{cyc} x^2y + \sum_{cyc} xy^2 &= 0
\end{align}
In the above equation, subtract $x^3+y^3+z^3$ from both sides.
\begin{align}
(x+y+z)^3 &= 1 \\
\sum_{cyc} x^3 + 3\sum_{cyc} x^2y + 3\sum_{cyc} xy^2 + 6xyz &= 1 \\
1 + 0 + 0 + 6xyz &= 1
\end{align}
Hence $xyz=0$.
