$(a-3)^3+(b-2)^3+(c-2)^3=0$, $a+b+c=2$, $a^2+b^2+c^2=6$. 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

There is a trouble with this question.
There are non-real solutions for example $a\approx 3.12128,\space b,c\approx -0.56064\pm 1.47835i$.
On the other hand, in $\mathbb R$ as in $\mathbb C$, if $a=2$ then the system 
$$-1+(y-2)^3+(z-2)^3=0\\y^2+z^2=2\\y+z=0$$ is incompatible  and if $b$ or $c$ are equal to $2$ then the system $$(x-3)^3+(y-2)^3=0\\x^2+y^2=2\\x+y=0$$ is incompatible too (i.e. none of the two systems has a solution).
Finally the conclusion is that there are no real numbers $a,b,c$ satisfying the three given equations
$$\begin{cases}(a-3)^3+(b-2)^3+(c-2)^3=0\\a+b+c=2\\a^2+b^2+c^2=6\end{cases}$$
Consequently there is no reason for this to imply that $a, b$ or $c$ must be equal to $2$.
(It could be argued that if the given system is incompatible then  $a, b$ or $c$ must be equal to $2$ so that the resulting system is also incompatible.This is not true by doing $a = 4$. I stop here without make $b$ or $c$ distinc of $2$).
