$(a-3)^3+(b-2)^3+(c-2)^3=0$, $a+b+c=2$, $a^2+b^2+c^2=6$. 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
 A: 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 $c=2$.
A: 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 $c=2$ 
A: 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 $2$).
A: solving the second equation for $c$ we get $$c=2-a-b$$ and inserting this in (I) and (III) we get after some Algebra
$$-9a^2+27a-35-6b^2+12b-3a^2b-3ab^2=0 (*)$$
$$2a^2+2b^2+2ab-4a-4b-2=0$$
solving the last equation for $b$:
$$b_{1,2}=-1/2\,a+1\pm 1/2\,\sqrt {-3\,{a}^{2}+4\,a+8}$$ inserting this in the  equation (*) we get
$$3\,{a}^{3}-9\,{a}^{2}+12\,a-41=0$$ after some Algebra, and the solution is given by $$a \approx3.091040052>2$$ 
the other case is analogously
