<s>[INST] $(a-3)^3+(b-2)^3+(c-2)^3=0$, $a+b+c=2$, $a^2+b^2+c^2=6$. Prove that at least one from $a,b,c$ is 2. 
Assume that $\{a,b,c\} \subset \Bbb R$, $(a-3)^3+(b-2)^3+(c-2)^3=0$, $a+b+c=2$, $a^2+b^2+c^2=6$.
Prove that at least one of the numbers $a, b, c\ $ is 2.

This is from a list of problems used for training a team for a math olympics. I tried to use known Newton identities and other symmetric polynomial results but without success (perhaps a wrong approach). Sorry if it is a duplicate. Hints and answers are always welcomed.
Edit: There is a problem with the original statement of the question in the original source. Under these assumptions it is impossible to have $a, b, c$ with value 2, as spotted in the comments and proved by the answers below
 [/INST] Suppose that $a=2$. Then $b+c=0$ and hence $b=-c$. The third equation gives $b^2+c^2=2$ hence $2b^2=2$ and $b^2=1$. If $b=1$ then $c=-1$ and vice versa. Then $(2-3)^3+(1-2)^3+(-1-2)^3=-1-1-27=-29$. The same thing happens when $b=-1$ and $c=1$. 
If $b=2$, then $a=-c$ again which leaves with $a$ being either $1$ or $-1$. Either way the first equation doesn't hold. Similar thing happens when $c=2$ 
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