Equilateral triangle $ABC$ with $O$ is the circumcenter and M is a point on $(O)$, is $MA^x+MB^x+MC^x$ a constant? 
Given an inscribed equilateral triangle $ABC$ with circumcenter $O$. M can be a point on any position of the circle $(O)$, except $A,B,C$. Prove or disprove that $MA^x+MB^x+MC^x$ is a constant for all integers $x$.


In a practice test, I got this question and it told me to prove that $MA=MB+MC$ (when chord $MA$ intersects chord $BC$), $MA^2+MB^2+MC^2=6R^2$ and $MA^4+MB^4+MC^4=18R^4$ (put $R=OA$).
I managed to solve them all (by putting an extra point $E$ on $AM$ so that $MB=ME$). However, with the use of computer (to draw the figure above) and calculator, I estimate that $MA^3+MB^3+MC^3=10R^3$; $MA^5+MB^5+MC^5=33R^5$; $MA^6+MB^6+MC^6=62R^6$; $MA^7+MB^7+MC^7=118R^7$.
Of course, that is just estimated, not exact because both the Sketchpad and the calculator don't give out the exact integer number. That's why I want to know if $MA^x+MB^x+MC^x$ is really a constant for all intergers $x$, calculated by $R$.

A necessary condition for the equality to hold in general is that the sums when $\,M =A\,$ and when $\,M=A'\,$ be equal, where $\,A'\,$ is the point diametrically opposed to $\,A\,$:
$$
\begin{align}
0^x + \left(R \sqrt{3}\right)^x + \left(R \sqrt{3}\right)^x = (2R)^x + R^x + R^x \;&\iff\; \left(2^x - 2 \left(\sqrt{3}\right)^x+2\right)R^x=0 \\
 &\iff\; 2^x - 2 \left(\sqrt{3}\right)^x+2=0 
\end{align}
$$
The latter equation has $\,x = 2, 4\,$ as the only two real roots.
Using properties of the centroid, $\,MA^2+MB^2+MC^2$ $= 3 MO^2 + OA^2+OB^2+OC^2$ $= 6 R^2\,$, so the sum is indeed constant for $\,x=2\,$.
For $\,x=4\,$ the sum is also constant, as follows from:


*

*$\,MA=MB+MC\,$ by Ptolemy's theorem for the cyclic quadrilateral $\,ABMC\,$

*by the law of cosines $\begin{cases}MB^2+MC^2+MB\,MC = 3R^2 \\ MA^2+MB^2-MA\,MB = 3R^2 \\ MA^2+MC^2 - MA\,MC = 3R^2 \end{cases}$, then multiplying the equalities by $\,MA^2, MC^2, MB^2\,$, respectively, and adding up:
$$\require{cancel} 2(MA^2MB^2+MB^2MC^2+MC^2MA^2) + \cancel{MA\,MB\,MC(MA-MB-MC)} = 3R^2(MA^2+MB^2+MC^2) = 18 R^4$$

*$MA^4+MB^4+MC^4 = \left(MA^2+MB^2+MC^2\right)^2 -2\left(MA^2MB^2+MB^2MC^2+MC^2MA^2\right) = 36R^4-18R^4=18R^4\,$
