I'm told that showing $a+b+c=x+y+z=0$ is enough. I don't actually see how to finish from there, but here's a stupidly funny proof of this claim. Let $d$ denote the common value of all six sums. Recall the amazing fact \[ a^2+b^2+c^2+(a+b+c)^2 = (a+b)^2+(b+c)^2+(c+a)^2. \]Now take the same expression for $x$, $y$, and $z$, and add the two equations together. Then \[ 3d + (a+b+c)^2 + (x+y+z)^2 = 3d; \]it follows that $a+b+c=x+y+z=0$. EDIT: Ah, a complex numbers approach finishes. (These can probably be used from the start, but eh.) The above yields that the complex numbers $r:=a+xi$, $s:=b+yi$, and $t:=c+zi$ have sum zero and have equal magnitudes; ergo, these three complex numbers are the vertices of an equilateral triangle centered at the origin. It follows that \[ 0 = \Re(r^2+s^2+t^2) = (a^2+b^2+c^2) - (x^2+y^2+z^2); \]the result follows.

Firstly, we see that $a=b=c=x=y=z=0$ works and that’s a solution. Now, suppose that there is a variable that isn’t $0$. Let’s suppose wlog that it’s $a$. Then, as all expressions are homogeneous, we can suppose wlog that $a=1$. Then, we need to see $2$ different cases: $\bullet$ If $x=0$ From the fact that $$b^2+y^2=(b+a)^2+y^2$$we see that $b=-\frac12$. Then, we check that $$a^2+x^2=b^2+y^2$$$$ \Leftrightarrow 1=\frac14+y^2$$$$ \Leftrightarrow y=\pm \frac{\sqrt3}2 $$ By symmetry, we see that $c=-\frac12$ and that $z=\pm \frac{\sqrt3}2$. Replacing this in the final expression, $$a^2+b^2+c^2=x^2+y^2+z^2$$$$ \Leftrightarrow 1+\frac14+\frac14=0+\frac34+\frac34$$$$ \Leftrightarrow \frac32=\frac32$$which is true. $\bullet$ If $x\neq0$ Firstly, we check that $$b^2+y^2=(a+b)^2+(x+y)^2$$$$ \Leftrightarrow 1+2b+x^2+2xy=0$$$$ \Leftrightarrow y=-\frac{1+2b+x^2}{2x}$$ Then, we calculate that $$a^2+x^2=b^2+y^2$$$$ \Leftrightarrow 1+x^2=b^2+\left(-\frac{1+2b+x^2}{2x}\right)^2$$$$\Leftrightarrow b^2+\frac{b^2}{x^2}+b+\frac{b}{x^2}+\frac1{4x^2}-\frac12-\frac{3x^2}4=0$$ We can check that we can divide it by $1+\frac1{x^2}$, giving $$b^2+b-\frac{3x^2}4+\frac14=0$$$$ \Leftrightarrow b=\frac{-1 \pm x\sqrt3}2$$ As $y=-\frac{1+2b+x^2}{2x}$, we obtain $y=-\frac{\pm\sqrt3+x}2$ By symmetry, we see that $c=\frac{-1 \pm x\sqrt3}2$ and $z=-\frac{\pm\sqrt3+x}2$ Obviously, the choice of the signal in $b$ implies the one in $y$ and the choice in $c$ implies the one in $z$. So, if we choose the same signal in $b$ and in $c$, it stays $$a^2+x^2=(b+c)^2+(x+y)^2$$$$ \Leftrightarrow 1+x^2=\left(-1 \pm x\sqrt3 \right)^2 + \left(\mp \sqrt3-x\right)^2$$$$\Leftrightarrow 1+x^2=4+4x^2 \Leftrightarrow x^2=-1$$which isn’t true for $x\in\mathbb{R}$. Then, they have different signs and let’s suppose wlog that $b$ has the $+$. Replacing it in $a^2+b^2+c^2=x^2+y^2+z^2$, it stays $$1+\left(\frac{-1+x\sqrt3}2\right)^2+\left(\frac{-1-x\sqrt3}2\right)^2=x^2+\left(-\frac{\sqrt3+x}2\right)^2+\left(-\frac{-\sqrt3+x}2\right)^2$$$$ \Leftrightarrow 1+\frac14+\frac{3x^2}4+\frac14+\frac{3x^2}4=x^2+\frac34+\frac{x^2}4+\frac34+\frac{x^2}4$$$$\Leftrightarrow \frac32+\frac{3x^2}2=\frac32+\frac{3x^2}2$$which ends the proof, QED.

Here is my 20 IQ entry for "worst solution" If the common sum is zero, then all numbers are zero. So WLOG the common quantity is $1$ by scaling. Write \begin{align*} 1 &= (a+b)^2 + \left( \pm \sqrt{1-a^2} \pm \sqrt{1-b^2} \right)^2 \\ &= a^2+2ab+b^2 + 2-(a^2+b^2) \pm 2\sqrt{(1-a^2)(1-b^2)} \\ \implies 2ab+1 &= 2\sqrt{(1-a^2)(1-b^2)} \\ \implies 4a^2b^2 + 4ab + 1 &= 4(1-a^2)(1-b^2) = 4 - 4a^2 - 4b^2 + 4(ab)^2 \\ \implies 4ab &= 4(1-a^2)(1-b^2) = 3 - 4a^2 - 4b^2 \\ \implies a^2+ab+b^2 &= \frac 34. \end{align*}Proceeding in the same way we get \[ b^2+bc+c^2 = \frac34, \quad c^2+ca+a^2 = \frac34. \]By using HMMT 2014 Problem A-9, this implies $ab+bc+ca$ is equal to some absolute constant $k$ (not depending on the choice of solution to the system). Hence, adding gives \[ a^2+b^2+c^2 = \frac 94 - \frac{1}{2} k. \]Similarly, $x^2+y^2+z^2 = \frac 94 - \frac{1}{2} k$, as needed.

If all the expressions equal $0$, every variable involved equal zero and the problem is done. Otherwise, WLOG scale them so all the expressions equal $1$. Let $ \alpha, \beta, \gamma$ be real numbers from the inverval $[0,2 \pi)$such that $$ a = sin(\alpha) \ \ x = cos(\alpha)$$$$ b = sin(\beta) \ \ y = cos(\beta)$$$$ c = sin(\gamma) \ \ z = cos(\gamma)$$ from $$a^2 + x^2 = (a+b)^2 + (x+y)^2$$we get $$ ab + xy = -1/2$$(remember variables got scaled such that the sum of squares equals 1) Then, $$ sin(\alpha) sin(\beta) + cos(\alpha) cos(\beta) = -1/2$$ which implies $$ cos(\alpha - \beta)= -1/2$$ and cyclically $$ cos(\beta - \gamma)= -1/2$$$$ cos(\gamma - \alpha)= -1/2$$ But the only options to $cos(x) = -1/2$ on$(-2 \pi, 2 \pi)$ are $\frac{-4}{3} \pi, \frac{-2}{3} \pi, \frac{2}{3} \pi, \frac{4}{3} \pi $ This, together with the fact that $\alpha - \beta + \beta - \gamma + \gamma - \alpha = 0$ forces $\alpha, \beta, \gamma$ to be: $$\theta -\frac{2}{3} \pi , \theta , \theta +\frac{2}{3} \pi$$in some order for some angle $\theta$. Finally,$$x^2 + y^2 +z^2 - a^2- b^2- c^2 $$$$= cos(2 \alpha) +cos(2 \beta) +cos(2 \gamma) $$$$= cos(\frac{4}{3} \pi + 2 \theta) +cos(2 \theta) +cos(\frac{-4}{3} \pi + 2 \theta) $$$$= \frac{-1}{2} cos(2 \theta) - \frac{\sqrt{3}}{2} sin(2 \theta) \ \ +cos(2 \theta) \ \ + \frac{-1}{2} cos(2 \theta) + \frac{\sqrt{3}}{2} sin(2 \theta) $$$$= 0 $$ And the problem is done

$a^2+x^2+b^2+y^2=2((a+b)^2+(x+y)^2)\implies a^2+b^2+x^2+y^2+4ab+4xy=0$, We do the same thing with $b,c,y,z$ and $c,a,z,x$. Add them up gives \[a^2+b^2+c^2+x^2+y^2+z^2+2ab+2bc+2ca+2xy+2yz+2zx=0\]So $(a+b+c)^2+(x+y+z)^2=0\implies a^2+b^2+c^2=x^2+y^2+z^2$ as desired.

Suppose no variable is zero. Then, notice that from $a^2+x^2=(a+b)^2+(x+y)^2$ we get that $|a+b|> |a| \iff |x|<|x+y|$. We can express this as $\mathrm{sign}(a)= \mathrm{sign}(b)$ iff $\mathrm{sign}(x)\neq\mathrm{sign}(y)$ which can be writen in a smart way as $$\mathrm{sign}(a)\mathrm{sign}(b)\mathrm{sign}(x)\mathrm{sign}(y)=-1$$Doing a cyclic product over $a,b,c$ and $x,y,z$ we get $$\big(\mathrm{sign}(a)\mathrm{sign}(b)\mathrm{sign}(c)\mathrm{sign}(x)\mathrm{sign}(y)\mathrm{sign}(z)\big)^2=-1$$A contradiction. Hence, one variable must be zero. After that the problem is easy to bash. Edit: I read this again after a while and I noticed that $\mathrm{sign}(a)\neq \mathrm{sign}(b)$ and $\mathrm{sign}(x)\neq\mathrm{sign}(y)$ simultaneously is perfectly possible so this solution is wrong lol.

Let $A = (a,x),$ $B=(b,y)$, $C=(c,z)$ The problem condition is $|A|=|B|=|C|=|A+B|=|B+C|=|C+A|$ From the identity $|A|^2 + |B|^2 = |A+B|^2 + |A-B|^2$, we get $|A-B| = |B-C| = |C-A|$. With the condition $|A|=|B|=|C|$, we see $A,B,C$ are points of equilateral triangle with center $(0,0)$... (*) (edited : an exceptional case is A,B,C are all the same points, but then |A|=|A+B|, so A,B,C must be (0,0)) Now define complex numbers $D = a+xi$, $E=b+yi$, $F=c+zi$ The goal is showing $a^2-x^2 + b^2-y^2 + c^2-z^2 = 0$, It is equivalent to show the sum of real part of $D^2 + E^2 + F^2$ is $0$. But by (*), $D,E,F$ is can be represented as $(p, wp, w^2p)$ where $w$ is the 3rd root of unity. We see $D^2 + E^2 + F^2$ is 0 so its real part is 0.

Sum up the last three and compare with the sum of the first three terms $(a+b)^2+(b+c)^2+(c+a)^2+(x+y)^2+(y+z)^2+(z+x)^2=(a^2+b^2+c^2)+(x^2+y^2+z^2)$ $(a+b+c)^2+(x+y+z)^2=0$ $a+b+c=x+y+z=0$ From $a^2+x^2=(a+b)^2+(x+y)^2$ >> $b^2+2ab=-(y^2+2xy)$ >> $(b^2+2ab)^2=(y^2+2xy)^2$ From $b^2+y^2=(a+b)^2+(x+y)^2$ >> $a^2+2ab=-(x^2+2xy)$ >> $(a^2+2ab)^2=(x^2+2xy)^2$ From $a^2+x^2=b^2+y^2$ >> $a^2-b^2=y^2-x^2$ >> $(a^2-b^2)^2=(y^2-x^2)^2$ Sum up all three equations $(a^2-b^2)^2+(a^2+2ab)^2+(b^2+2ab)^2=(y^2-x^2)^2+(x^2+2xy)^2+(y^2+2xy)^2$ $2a^4+4a^3b+6a^2b^2+4ab^3+2b^4=2x^4+4x^3y+6x^2y^2+4xy^3+2y^4$ $2(a^2+ab+b^2)^2=2(x^2+xy+y^2)^2$ Since $a^2+ab+b^2=(a+\frac{b}{2})^2+\frac{3b^2}{4}\geq0$ Similarly, $x^2+xy+y^2\geq 0$ Hence, $a^2+ab+b^2=x^2+xy+y^2$ So, $a^2+b^2+c^2=a^2+b^2+(-a-b)^2=2(a^2+ab+b^2)=2(x^2+xy+y^2)=x^2+y^2+(-x-y)^2=x^2+y^2+z^2$ as desired.

Notice that by condition we have \[a^2+x^2 = a^2+2ab+b^2 + x^2+2xy+y^2\implies -2(ab+xy) = b^2+y^2 = k\]where let $k$ denote the common value of the six terms. By symmetry this gives $2(ab+bc+ca)+2(xy+yz+zx) = -3k$. However on the other hand, by comparing the sum of the first three expressions with that of the last three expressions, we have that $2(ab+bc+ca)+2(xy+yz+zx) = 0$. Thus we have $k = 0$. Notice that this implies $a = b = c = x = y = z = 0$, which clearly implies the result.

Let $t$ the common value $a^2+x^2=b^2+y^2=c^2+z^2={(a+b)}^2+{(x+y)}^2={(b+c)}^2+{(y+z)}^2={(c+a)}^2+{(z+x)}^2. $ So $ 2t = a^2 + b^2 + x^2 + y^2$ and $t = a^2 + b^2 + x^2 + y^2 + 2ab + 2xy $ and so $2ab + 2xy = -t $ and so $c^2 + z^2 + 2ab + 2xy = 0$ and similarly $a^2 + x^2 + 2bc + 2yz = b^2 + y^2 + 2ac + 2xz = 0$ and suming we get ${(a+b+c)}^2 + {(x+y+z)}^2 = 0$ Thus $a + b + c = x + y + z = 0.$ Then let $p,q$ be complex numbers with $ p = a+ xi$ and $q = b + yi$ and consequently $p + q = -c - zi$ and we know $|p| = |q| = |p + q|$ and we can simplify this to be $(p + q)\left(\dfrac{1}{p} + \dfrac{1}{q} \right) = 1$ which implies ${\left(\dfrac{q}{p} \right)}^2 + \dfrac{q}{p} +1 = 0$ and $\dfrac{q}{p} = \dfrac{-1}{2} + i\dfrac{\sqrt{3}}{2}$ or $\dfrac{p}{q} = \dfrac{-1}{2} + i\dfrac{\sqrt{3}}{2}.$ SWLOG $\dfrac{q}{p} = \dfrac{-1}{2} + i\dfrac{\sqrt{3}}{2}$ and hence $ b + yi = (a+ xi)\left(\dfrac{-1}{2} + i\dfrac{\sqrt{3}}{2} \right) = \left(-\dfrac{a}{2} -\dfrac{x\sqrt{3}}{2}\right) +i\left(\dfrac{a\sqrt{3}}{2} - \dfrac{x}{2} \right)$ and hence $ b = -\dfrac{a}{2} -\dfrac{x\sqrt{3}}{2} $ $ y = \dfrac{a\sqrt{3}}{2} - \dfrac{x}{2} $ and so $c = -a -b = -\dfrac{a}{2} +\dfrac{x\sqrt{3}}{2}$ $z = -x - y = -\dfrac{a\sqrt{3}}{2} - \dfrac{x}{2} $ and we can easily conclude $a ^2 + b^2 + c^2 = x^2 + y^2 + z^2 = \dfrac{3}{2} (a^2 + x^2)$

Solved with brain damage. Label the six expressions in the display $(1), (2)$, etc. Notice that summing $(4)+(5)+(6)-(1)-(2)-(3)$ yields \begin{align*} x^2+y^2+z^2+a^2+b^2+c^2+2xy+2xz+2yz+2ab+2bc+2ca&=0\\ (x+y+z)^2+(a+b+c)^2 &= 0. \end{align*}As a result $x+y+z=0$ and $a+b+c=0$. Then we have \begin{align*} b^2+y^2 &= a^2+b^2+x^2+y^2+2xy+2ab \iff a^2+2ab+x^2+2xy = 0 \\ a^2+x^2 &= a^2+b^2+x^2+y^2+2xy+2ab \iff b^2+2ab+y^2+2xy = 0. \end{align*}Subtracting the two equations and rearranging, $$a^2-b^2=x^2-y^2 \iff a^2+y^2 = b^2+x^2.$$But combined with the initial conditions, this yields $x^2=y^2=(x+y)^2$ by symmetry. In both the $x=y$ and $x=-y$ case, we obtain $x=y=z=0$; similarly, $a=b=c=0$. Thus $x=y=z=a=b=c=0$ is the only solution, which clearly satisfies the desired equation.

HamstPan38825 wrote: Then we have \begin{align*} b^2+y^2 &= a^2+b^2+x^2+y^2+2xy+2ab \iff a^2+2ab+x^2+2xy = 0 \\ a^2+x^2 &= a^2+b^2+x^2+y^2+2xy+2ab \iff b^2+2ab+y^2+2xy = 0. \end{align*}Subtracting the two equations and rearranging, $$a^2-b^2=x^2-y^2 \iff a^2+y^2 = b^2+x^2.$$ Is this part correct? I'm not entirely sure if I'm missing something or there's a sign error.

EDIT: wait darn i'm like really stupid pls ignore this solution i'll try to fix soon

isn't necessary anyways.

Remarks: What a problem.

Subtracting the sum of the first three from the sum of the last three we get: $$(a+b+c)^2=-(x+y+z)^2\iff a+b+c=x+y+z=0.$$Let all of these $6$ equal $M.$ In particular we have: $$2(ab+xy+M)=a^2+b^2+x^2+y^2+2ab+2xy=M.$$Substituting here $x=\sqrt{M-a^2}$ and $y=\sqrt{M-b^2}$ and solving for $M$ we get: $$\frac 43(a^2+ab+b^2)=M=(a+b)^2+(x+y)^2\iff a^2+ab+b^2=x^2+xy+y^2.$$Now we have: \begin{align*} &a^2+b^2+c^2=a^2+b^2+(-a-b)^2=2(a^2+ab+b^2) \\ =& \; 2(x^2+xy+y^2)=x^2+y^2+(-x-y)^2=x^2+y^2+z^2\end{align*}

Summing this cyclicly, we get $0=(a+b+c)^2+(x+y+z)^2$, i.e. $x+y+z=0$. But notice that from $b^2+y^2=(a+b)^2+(x+y)^2$ we also get $0=2ab+a^2+2xy+x^2$, and subtracting this from (\ref{1}) gives $a^2+y^2=b^2+x^2$. Together with the first equation, this implies $x^2=y^2$ (and $a^2=b^2$). Similarly, we get $y^2=z^2$ (and $b^2=c^2$), i.e. $|x|=|y|=|z|$, which together with $x+y+z=0$ implies $x=y=z=0$ (and $a=b=c=0$). From here the result is obvious.

To prove this statement, we can subtract $(a+b)^2+(x+y)^2$ from both sides of the equation $(c+a)^2+(z+x)^2 = (a+b)^2+(x+y)^2$. This gives us \[(c+a)^2+(z+x)^2 - (a+b)^2-(x+y)^2 = 0 \] Expanding the left side of this equation and rearranging the terms gives us \[2c^2 - 2a^2 + 2z^2 - 2x^2 = 0 \] Similarly, we can subtract $(b+c)^2+(y+z)^2$ from both sides of the equation $(c+a)^2+(z+x)^2 = (b+c)^2+(y+z)^2$. This gives us \[(c+a)^2+(z+x)^2 - (b+c)^2-(y+z)^2 = 0 \] Expanding the left side of this equation and rearranging the terms gives us \[2a^2 - 2b^2 + 2x^2 - 2y^2 = 0 \] Adding these two equations gives us \[2c^2 - 2a^2 + 2z^2 - 2x^2 + 2a^2 - 2b^2 + 2x^2 - 2y^2 = 0 \] Simplifying this equation gives us \[2c^2 - 2b^2 + 2z^2 - 2y^2 = 0 \] Adding this equation to the previous one gives us \[2c^2 - 2a^2 + 2z^2 - 2x^2 + 2a^2 - 2b^2 + 2x^2 - 2y^2 + 2c^2 - 2b^2 + 2z^2 - 2y^2 = 0 \] Simplifying this equation gives us \[4c^2 + 4z^2 - 4a^2 - 4x^2 - 4b^2 - 4y^2 = 0 \] This simplifies to \[(c^2 + z^2) - (a^2 + x^2) - (b^2 + y^2) = 0 \] But we are given that $a^2 + x^2 = b^2 + y^2 = c^2 + z^2$, so this equation becomes \[(c^2 + z^2) - (c^2 + z^2) - (c^2 + z^2) = 0 \] Simplifying this equation gives us \[-3c^2 - 3z^2 = 0 \] This simplifies to \[-c^2 - z^2 = 0 \] Adding this equation to the equation $2c^2 - 2a^2 + 2z^2 - 2x^2 = 0$ gives us \[c^2 + z^2 - a^2 - x^2 = 0 \] This simplifies to \[c^2 - a^2 + z^2 - x^2 = 0 \] Starting from the equation $c^2 - a^2 + z^2 - x^2 = 0$, we can add $(b+y)^2 - (a+x)^2$ to both sides to get \[(c+z)^2 - (a+x)^2 + (b+y)^2 - (a+x)^2 = 0 \] This simplifies to \[(c+z)^2 + (b+y)^2 - 2(a+x)^2 = 0 \] Since $(c+z)^2 + (b+y)^2$ and $(a+x)^2$ are both positive, we can divide both sides of the equation by $(a+x)^2$ to get \[\frac{(c+z)^2 + (b+y)^2}{(a+x)^2} - 2 = 0 \] This simplifies to \[\frac{(c+z)^2 + (b+y)^2}{(a+x)^2} = 2 \] Since the left side of this equation is a ratio of two squares, it must be positive. Therefore, we must have $(c+z)^2 + (b+y)^2 > 0$ and $(a+x)^2 > 0$. Since all of the variables are real numbers, we must have $c+z \ne 0$ and $b+y \ne 0$. Therefore, we can divide both sides of the equation by $(c+z)^2 + (b+y)^2$ to get \[\frac{(a+x)^2}{(c+z)^2 + (b+y)^2} = \frac{1}{2} \] This equation can be written as \[(a+x)^2 = \frac{1}{2}((c+z)^2 + (b+y)^2) \] Since $(a+x)^2$, $(c+z)^2$, and $(b+y)^2$ are all nonnegative, we must have \[(a+x)^2 \le \frac{1}{2}((c+z)^2 + (b+y)^2) \] with equality if and only if $(a+x)^2 = 0$. But we are given that $(a+x)^2 = (b+y)^2 = (c+z)^2$, so we must have \[(a+x)^2 = \frac{1}{2}((c+z)^2 + (b+y)^2) = (b+y)^2 = (c+z)^2 \] This means that $(a+x)^2 = (b+y)^2 = (c+z)^2 = 0$, which implies that $a+x = b+y = c+z = 0$. Substituting $a+x=0$, $b+y=0$, and $c+z=0$ into the equation $a^2+b^2+c^2=x^2+y^2+z^2$ gives us \[(-x)^2 + (-y)^2 + (-z)^2 = x^2 + y^2 + z^2 \] This equation simplifies to \[-2x^2 - 2y^2 - 2z^2 = 0 \] Since $-2x^2$, $-2y^2$, and $-2z^2$ are all nonpositive, we must have \[-2x^2 - 2y^2 - 2z^2 \le 0 \] with equality if and only if $-2x^2 = -2y^2 = -2z^2 = 0$. But we are given that $-2x^2 = -2y^2 = -2z^2$, so we must have \[-2x^2 - 2y^2 - 2z^2 = -2x^2 - 2y^2 - 2z^2 = 0 \] This means that $-2x^2 = -2y^2 = -2z^2 = 0$, which implies that $x=y=z=0$. Therefore, we must have $a=b=c=x=y=z=0$, which means that all of the variables are equal to $0$. This satisfies the original equation, so the proof is complete.

Note the problem is homogeneous so WLOG let $a^2+x^2=1$. Then, let $a=\sin\alpha$, $x=\cos\alpha$, $b=\sin\beta$, $y=\cos\beta$, $c=\sin\gamma$, $z=\cos\gamma$. WLOG let $0\le \gamma\le \beta\le \alpha\le 2\pi$. We see $(a+b)^2+(x+y)^2=1$ reduces to $\cos(\alpha-\beta)=-1/2$ so $\alpha-\beta=2\pi/3,4\pi/3$. Similarly, $\beta-\gamma=2\pi/3,4\pi/3$ and $\gamma-\alpha=-2\pi/3,-4\pi/3$. Hence, we see that $(\alpha,\beta,\gamma)=(\gamma+4\pi/3,\gamma+2\pi/3,\gamma)$. We can check that $\sum_{\text{cyc}}(a^2-x^2)=\sum_{\text{cyc}}\cos(2\alpha)$ is equal to zero. $\square$

Label the expressions [1],[2],...,[6]. $$[6]+[5]+[4]-[3]-[2]-[1]\implies (a+b+c)^2+(x+y+z)^2=0\iff a+b+c=0=x+y+z.$$Now, interpret this as complex numbers $d=a+xi$ etc. with sum 0 and equal magnitudes, which means they form an equilateral triangle with center at the origin. Hence $a^2-x^2+b^2-y^2+c^2-z^2=d^2+e^2+f^2=0$ which is well known (say, by Vieta's on $x^3-1=0$ which the roots represent the vertices, up to scaling of magnitude). $\blacksquare$

The trig solution is much more natural, but this is cooler (solved with ARCH hint) Adding the first 3 sums and subtracting the last 3 yields $$(a+b+c)^2+(x+y+z)^2=0 \Rightarrow a+b+c=x+y+z=0.$$Now, let $z_1=a+xi$, $z_2=b+yi$, $z_3=c+zi$. Notice that $|z_1|=|z_2|=|z_3|$ and $|z_1+z_2+z_3|^2=(a+b+c)^2+(x+y+z)^2=0$, which means that the triangle is equilateral. Then, we have $\textrm{Re}(z_1^2+z_2^2+z_3^2)=0$. But by expansion, we also know that $$\textrm{Re}(z_1^2+z_2^2+z_3^2)=0=(a^2+b^2+c^2)-(x^2+y^2+z^2) \Rightarrow a^2+b^2+c^2=x^2+y^2+z^2,$$as desired.

Scale so the sums equal $1$. Let $z_1=a+xi$, $z_2=b+yi$, $z_3=c+zi$. Then, we have \[|z_1|=|z_2|=|z_3|=|z_1+z_2|=|z_2+z_3|=|z_3+z_1|=1\] implying that $z_1, z_2, z_3$ are the vertices of an equilateral triangle. Hence, \[z_1^2+z_2^2+z_3^2=0\] so the real parts are equivalent, or \[a^2+b^2+c^2-x^2-y^2-z^2=0\] implying our desired conclusion. $\square$

Note that $a=b=c=x=y=z=0$ is a valid solution which also satisfies the condition we wish to prove. Suppose $a$, $b$, and $c$ are not all distinct, or WLOG $a=b$. The only way this is possible is if \[a=b=c=x=y=z=0,\] which satisfies the condition we wish to prove. From here, we can assume $a,b,c$ are distinct and $x,y,z$ are distinct. The sum of the first 3 expressions is equal to the sum of the last 3 expressions, so \[(a+b+c)^2 + (x+y+z)^2 = 0\]\[\implies a+b+c = x+y+z = 0.\] Define the complex numbers $\omega_1 = a+xi$, $\omega_2 = b+yi$, and $\omega_3 = c+zi$. Then we have \[|\omega_1| = |\omega_2| = |\omega_2|, \qquad \omega_1 + \omega_2 + \omega_3 = 0. \] Since we taken care of the case when $\omega_1 = \omega_2 = \omega_3 = 0$, we can scale them such they are on the unit circle. Using vectors, it follows that $\omega_1$, $\omega_2$, and $\omega_3$ form an equilateral triangle. Hence \[0 = \operatorname{Re}\left(\omega_1^2 + \omega_2^2 + \omega_3^2\right) = \left(a^2+b^2+c^2\right) - \left(x^2+y^2+z^2\right).~\blacksquare\]

Consider complex numbers $z_1, z_2, z_3$ such that $z_1 = a + xi, z_2 = b + yi, z_3 = c + z_i$. Then the given conditions imply that $|z_1| = |z_2| = |z_3| = |z_1 + z_2| = |z_2 + z_3| + |z_3 + z_1| = r$ for some nonnegative $r$. By considering the circle centered at $z_1$ in the complex plane with radius $r$, we find that $z_1, z_2, z_3$ form the vertices of an equilateral triangle. Thus, for some primitive third root of unity $\omega$, we have $z_2 = \omega z_1$ and $z_3 = \omega^2z_1$. Therefore $$\begin{aligned} a^2 - x^2 + b^2 - y^2 + c^2 - z^2 &= \text{Re}(z_1^2 + z_2^2 + z_3^2) \\ &= \text{Re}(z_1^2(1 + \omega^2 + \omega^4)) \\ &= \text{Re}(z_1^2)(0) \\ &= 0,\end{aligned}$$so we are done.

Since $a, b, c, x, y, z$ are real, define $r = a + xi, s = b + yi, t = c + zi$. The equation provided is equivalent to $$|r| = |s| = |t| = |r + s| = |s + t| = |t + r|.$$Observe that $r = s = t = 0$ is a valid solution to the problem, we get $a = b = c = x = y = z = 0$ from this which satisfies $a^2 + b^2 + c^2 = x^2 + y^2 = z^2$. Now suppose the magnitude is nonzero. Without loss of generality, assume $|r| = 1$. If $|k| = 1$, then if $(r, s, t)$ is a valid solution if and only if $(rk, sk, tk)$ is also valid, hence, again assume without loss of generality that $r = 1$. Now our equation becomes $$1 = |s| = |t| = |1 + s| = |s + t| = |1 + t|.$$It is well-known that the only solutions for $s, t$ is some permutation of $\left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i, -\frac{1}{2} + \frac{\sqrt{3}}{2}i \right)$. This implies that $r, s, t$ form an equilateral triangle, so $r^2 + s^2 + t^2 = 0$. Then $$0 = \Re(r^2 + s^2 + t^2) = \Re(r^2) + \Re(s^2) + \Re(t^2) = (a^2 - x^2) + (b^2 - y^2) + (c^2 - z^2).$$We have shown $a^2 + b^2 + c^2 = x^2 + y^2 + z^2$. Our proof is complete.

Let $\zeta_1 = a + xi$, $\zeta_2 = b + yi$, and $\zeta_3 = c + zi$. Since the equations are homogeneous, we can assume that $\newline$ \[|\zeta_1| = |\zeta_2| = |\zeta_3| = |\zeta_1 + \zeta_2| = |\zeta_2 + \zeta_3| = |\zeta_3 + \zeta_1| = 1\]Which implies that $\zeta_1$, $\zeta_2$, and $\zeta_3$ are the vertices of an equilateral triangle on the unit circle. $\newline$ And since the triangle is a rotation of the $3$rd roots of unity, we have \[\{a, b, c\} = \{x, y, z\} \implies a^2 + b^2 + c^2 = x^2 + y^2 + z^2.\]

Notice $a=b=c=x=y=z=0$ succeeds; now presume otherwise. We will now scale down the problem to where $a^2+x^2=1$. Denote $p=a+xi$, similarly defining $q$ and $r$. Our condition implies $p$, $q$, $r$, $p+q$, $q+r$, $r+s$ all lie on the unit circle. Hence, $p$, $q$, $r$ are some rotation of the third roots of unity. As a somewhat obscure property of the third roots of unity and their rotations, $\sum\Re(p)^2=\sum\Im(p)^2$.

Focus on $a^2 + x^2 = b^2 + y^2 = (a+b)^2 + (x+y)^2.$ We can subtract this equation from $a^2 + x^2 + b^2 + y^2 = a^2 + x^2 + b^2 + y^2 = a^2 + x^2 + b^2 + y^2$ to get \[ a^2 + x^2 = b^2 + y^2 = -2(ab+xy). \]Therefore, our given relation can be written as \[ a^2 + x^2 = b^2 + y^2 = c^2 + z^2 = -2(ab+xy)=-2(bc+yz)=-2(ca+zx). \]Now, in vector notation, let $u = \binom{a}{x}, v = \binom{b}{y},$ and $z = \binom{c}{z}.$ Then the given relation translates into \[ |u|^2 = |v|^2 = |z|^2 = -2(u \cdot v) = -2(v \cdot w) = -2(w \cdot u), \]where $\cdot$ denotes the dot product. This implies that the cosine of the angle between the vectors $u$ and $v$ is $-\frac{1}{2}.$ Therefore, the vectors $u,v,w$ have the same magnitude and are spaced from each other by $120^\circ$. Now, let $p,q,r$ be the complex numbers representing the vectors $u,v,w$, respectively. Then $q = \omega p$ and $r = \omega^2 p$ where $\omega$ is a third root of unity. Thus \[ p^2 + q^2 + r^2 = p^2 (1 + \omega^2 + \omega^4) = 0. \]On the other hand, we see that $p^2 = (a+ix)^2 = (a^2 - x^2) + (2ax)i,$ and similarly for $q$ and $r.$ Thus \[ (a^2+b^2+c^2-x^2-y^2-z^2) + (2ax+2by+2cz)i = 0. \]Comparing real parts gives the desired result.

Note that all of the expressions are degree $2$, so we can scale variables without affecting the equalities, as long as they aren't all equal to $0$. However, the $0$ case is trivial, since then all of the squares are $0$ and clearly $0=0$. WLOG assume that all of the expressions equal $1$. Then we let $a=\cos{p}$, $x=\sin{p}$, $b=\cos{q}$, $y=\sin{q}$, $c=\cos{r}$, $z=\sin{r}$ to satisfy the first two equalities. Then we have that $2+2\cos{p}\cos{q}+2\sin{p}\sin{q}=1$. By trig identities, the left side simplifies to $2+2\cos{p-q}$, so we end up with $\cos{p-q}=-\frac12$, and $p-q=\pm \frac{2\pi}{3}$. We similarly get that $q-r\pm \frac{2\pi}{3}$ and $r-p=\pm \frac{2\pi}{3}$, which altogether tells us that the three angles $p$, $q$, and $r$ must be evenly spaced around the unit circle. To finish, note that the three equally spaced points on the unit circle $(a,x)$, $(b,y)$, and $(c,z)$ are solutions to the equation $x^3=u$ for some constant complex number $u$ on the unit circle. By Vieta's, the sum of the squares of the roots is $0$. This sum is also equivalent to the sum of the squares of the real parts of the roots, minus the sum of the squares of the imaginary parts, which tells us that $a^2+b^2+c^2-x^2-y^2-z^2=0$, as desired.

Groupsolved with thoomgus. I used a complex numbers solution, but frankly, I don't think we need another one. Here's something more synthetic that thoom came up with: Consider the complex plane, and denote $O = 0$, $A = a+xi$, $B = b+yi$, $C = c+zi$. Now, denote $P = B + C$, $Q = A + C$, $R = A + B$. We may now use some parallelogram properties. Note that $P+Q$ must also lie on the circle, as well as forming a parallelogram $O,A,B,R$. Now, we have $\overline{OA} = \overline{BR} = \overline{OB} = \overline{AR}$, implying that $O,A,B,R$ is a rhombus with $\angle ROQ = 120^{\circ}$. Repeating this for $B,C$ means that $ABC$ and $PQR$ are equilateral triangles, and the result is obvious.

this solution is the reason downvotes should be brought back :skull: first paragraph is probably unnecessary as well but whatever Summing everything yields that $$a^2+b^2+c^2+x^2+y^2+z^2=(a+b)^2+(b+c)^2+(c+a)^2+(x+y)^2+(y+z)^2+(x+z)^2.$$However, note that $\sum(a+b)^2-\sum a^2=(a+b+c)^2$ for all $(a,b,c)$ so we have that $(a+b+c)^2+(x+y+z)^2=0$ which is enough to show that $a+b+c=x+y+z=0.$ Now, scale the equations such that the common sum is $1$. Also, let $a=\sin{a}, b=\sin b, c=\sin c, x=\cos a, y=\cos b, z=\cos c$. Then, the last three equations give us that $$\sin^2{a}+\sin^2b+\cos^2a+\cos^2b+2(\sin a \sin b + \cos a \cos b)=1 \implies \cos (b-a)=-\frac{1}{2}$$and their analogous variants. This, along with $a+b+c=x+y+z=0$, is enough to imply that $a,b,c$ can be rewritten as $x-\frac{2\pi}{3}, x, x+\frac{2\pi}{3}$ in some order. Then, the condition that $x^2+y^2+z^2=a^2+b^2+c^2$ is equivalent to proving that $\cos(2x-\frac{4\pi}{3})+\cos(2x)+\cos(2x+\frac{4\pi}{3})=0$ by double angle formula. However, from the sum to product identity on $\cos(2x-\frac{4}{3})+\cos(2x+\frac{4\pi}{3})=-\cos(2x)$ so we want $\cos(2x)-\cos(2x)=0$, which is obvious so we are done.