Let $\displaystyle \omega = e^{2\pi i/3}$ be a complex cube root of unity. Then

$\displaystyle (a+b+c)^2 = a^2 +b^2 +c^2 +2bc+2ca+2ab = x+y+z,$

$\displaystyle (a+\omega b+\omega^2c)^2 = a^2 +\omega^2b^2 +\omega c^2 +2bc+2\omega^2ca+2\omega ab = x+\omega^2y+\omega z,$

$\displaystyle (a+\omega^2 b+\omega c)^2 = a^2 +\omega b^2 +\omega^2 c^2 +2bc+2\omega ca+2\omega^2 ab = x+\omega y+\omega^2 z.$

Let $\displaystyle X$ be one of the two square roots of $\displaystyle x+y+z$, and similarly let $\displaystyle Y = \bigl(x+\omega y+\omega^2 z\bigr)^{1/2}$ and $\displaystyle Z = \bigl(x+\omega^2 y+\omega z\bigr)^{1/2}.$ Thus there are eight possible choices for the triple of complex numbers $\displaystyle (X,Y,Z).$

Then $\displaystyle a+b+c=X$, $\displaystyle a+\omega^2 b+\omega c = Y$ and $\displaystyle a+\omega b+\omega^2c = Z.$

Since $\displaystyle 1+\omega+\omega^2=0$, it follows that $\displaystyle X+Y+Z = 3a$, $\displaystyle X+\omega Y + \omega^2 Z = 3b$ and $\displaystyle X+\omega^2 Y + \omega Z = 3c.$

That gives the solution in the neat form

$\displaystyle a = \frac13(X+Y+Z),$

$\displaystyle b = \frac13(X+\omega Y + \omega^2 Z),$

$\displaystyle c = \frac13(X+\omega^2 Y + \omega Z).$

Of course, if you write X, Y and Z in terms of x, y and z, it won't look quite so neat. But I reckon that I can claim

**Opalg: 1, Mathematica: 0.**