General solutions of $x+y+z=x^2+y^2+z^2=x^3+y^3+z^3$ Note: this actually has nothing to do with probability theory. Fell free to skip ahead and solve the equations below. 
Since I am unable to solve this continuous version of a problem about moments, I am now trying to do something simple but related. Let's say $X$ is a random variable which can take three values $x,y,z$ with probability $1/3$ each. For what values of $x,y,z$ does $X$ have its first, second and third central moments equal?
That is, how to solve the system of equations
$$
x+y+z=x^2+y^2+z^2=x^3+y^3+z^3?
$$
The first two equations give a sphere which just covers the unit cube $0<|x|,|y|,|z|<1$. But the second equation is harder to tackle. I guess that we can only have trivial solutions at $x,y,z=0,1$, but I am not sure. Writing $p=x+y+z$, we get a cubic with solution $x,y,z$:
$$
u^3-pu^2+\frac12 p(p-1)u-\frac{1}{12}p(p-1)(p-2) =0,
$$
but that doesn't seem to help.

Let the values be equal to $t$.   
Then $xy+yz+zx = \frac{ (x+y+z)^2 - (x^2+y^2+z^2) }{2} = \frac{t^2-t}{2}$
and $xyz = \frac{ (x+y+z)^3 - (x^3+y^3+z^3) - 3(xy+yz+zx)(x+y+z)}{-3} = \frac{t^3-t - 3(t^2-t)/2 \times t}{-3} = \frac{-t^3+3t^2 - 2t } { 6}$.
So, these are roots to the cubic
$$X^3 - 6tX^2 + 3(t^2-t)X+(t^3-3t^2+2t) = 0.$$
We could then obtain the solutions via the cubic formula:

