<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] Let me try. Note that $2(ab+bc+ca) = a^2+b^2+c^2-(a+b+c)^2 = 2$, then $ab+bc+ca=1$.
We have $$(a-3)^3 + (b-2)^3+(c-2)^3 = 0$$
$$(a-3)^3+(b+c-4)(b^2+c^2+4-2b-2c-bc) = 0$$
$$(a-3)^3 + (-2-a)(6-a^2+4-2(2-a)-(1-a(2-a)))=0$$
$$(a-3)^3-(2+a)(-2a^2+4a+5)=0$$
$$3a^3-9a^2+14a-37=0$$
$$3(a-1)^3+5(a-1)-29=0$$
Solve this equation, you get $a = 2$ is not the root. 
If $b=2$, then $a+c=0$, $a^2+c^2 = 2$, so $a=1$, $c=-1$ but it doesn't satisfy the first equation. 
Similarly for $c=2$.
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