<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] 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.
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