<s>[INST] Given six numbers $x,y,z,a,b,c$ which satisfy the following relations. Express $x+y+z$ in terms of $a,b,c$ Given six numbers $x,y,z,a,b,c$ which satisfy the following relations
$y^2+yz+z^2=a^2$
$z^2+zx+x^2=b^2$
$x^2+xy+y^2=c^2$
Express $x+y+z$ in terms of $a,b,c$  
My attempt:
$\dfrac{(y^3-z^3)}{(y-z)}=a^2,\dfrac{(z^3-x^3)}{(z-x)}=b^2,\dfrac{(x^3-y^3)}{(x-y)}=c^2$
But this step didn't seem to help.
So I tried adding and expressing $k(a+b+c)^2$ but some other unmanagable elements came along. I could not think of any other way to approach the problem. Please help.
 [/INST] Let $A=a^2,B=b^2,C=c^2$. We have 
$$
\begin{array}{lcl}
B-A &=& (x-y)(x+y+z) \\
C-A &=& (x-z)(x+y+z) \\
C-B &=& (y-z)(x+y+z) \\
\end{array}\tag{1}
$$
Then, if we put $D=(B-A)^2+(C-A)^2+(C-B)^2$, we deduce
$$
\begin{array}
D
&=& (x+y+z)^2\Bigg((x-y)^2+(x-z)^2+(y-z)^2\Bigg) \\
&=& 2(x+y+z)^2\Bigg(x^2+y^2+z^2-xy-xz-yz\Bigg) \\
&=& 2(x+y+z)^2\Bigg(
\big(2(x^2+y^2+z^2)+xy+xz+yz\big)-\big(x^2+y^2+z^2+2(xy+xz+yz)\big)
\Bigg) \\
&=& 2(x+y+z)^2\Bigg(
\big(A+B+C\big)-\big((x+y+z)^2\big)
\Bigg) \\
\end{array}\tag{2}
$$
So the sum $w=x+y+z$ satisfies 
$$
2w^4-2(A+B+C)w^2+(B-A)^2+(C-A)^2+(C-B)^2=0 \tag{3}
$$
If we simplify by 2 we obtain
$$
w^4-(a^2+b^2+c^2)w^2+(a^4+b^4+c^4-((ab)^2+(ac)^2+(bc)^2)=0 \tag{4}
$$
</s>