<s>[INST] Can we conclude that two of the variables must be $0$? 
Assuming $$a^2+b^2+c^2=1$$ and $$a^3+b^3+c^3=1$$ for real numbers $a,b,c$, can we conclude that two of the numbers $a,b,c$ must be $0$ ?

I wonder whether mathworld's result that only the triples $(1,0,0)$ , $(0,1,0)$ , $(0,0,1)$ satisfy the given equation-system , is actually true.
Looking at $(a+b+c)^3$ and $(a+b+c)^2$ , using 
\begin{align}
&(a+b+c)^3= (a+b+c)^2(a+b+c)=\\  
&(1+2(ab+ac+bc))(a+b+c)= \\
&2(a+b)(a+c)(b+c)+a+b+c+2abc
\end{align}
and eliminating $(a+b)(a+c)(b+c)$, with $S:=a+b+c$ , I finally got  $$(S-1)^2(S+2)=6abc$$
I guess this is not enough to show the above result (if it is true at all).
This question is inspired by an exercise to determine the possible values of $a+b+c$ assuming the above equations, so this question could be a duplicate, but I am not sure whether it actually is.
 [/INST] Since $|a|,|b|,|c|\in[0,1]$ (because $a^2+b^2+c^2=1$), we have
$$
a^2(1-a)+b^2(1-b)+c^2(1-c)=0
$$
The LHS is a sum of non-negative terms; to be zero, all the terms must be zero. This means that
$$
a^2(1-a)=b^2(1-b)=c^2(1-c)=0
$$ 
and therefore $|a|,|b|,|c|\in\{0,1\}$. Along with $a^2+b^2+c^2=1$, we get that exactly one of them is $1$, and two must be zero: this implies the result.
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