How to solve this system of equation. $x^2-yz=a^2$
$y^2-zx=b^2$
$z^2-xy=c^2$
How to solve this equation for $x,y,z$. Use elementary methods to solve (elimination, substitution etc.).
Given answer is:$x=\pm\dfrac{a^4-b^2c^2}{\sqrt {a^6+b^6+c^6-3a^2b^2c^2}}\,$, $y=\pm\dfrac{b^4-a^2c^2}{\sqrt {a^6+b^6+c^6-3a^2b^2c^2}}\,$ ,$z=\pm\dfrac{c^4-b^2a^2}{\sqrt {a^6+b^6+c^6-3a^2b^2c^2}}$
 A: Observe that $$(a^2)^2-b^2\cdot c^2=(x^2-yz)^2-(y^2-zx)(z^2-xy)=x(x^3+y^3+z^3-3xyz)$$
Similarly, $$b^4-c^2a^2=y(x^3+y^3+z^3-3xyz)\text{ and }c^4-a^2b^2=z(x^3+y^3+z^3-3xyz)$$
So, $$\frac x{a^4-b^2c^2}=\frac y{b^4-c^2a^2}=\frac z{c^4-a^2b^2}=\frac1{x^3+y^3+z^3-3xyz}=k\text{(say)}$$
Now, $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=(x+y+z)\frac{\{(x-y)^2+(y-z)^2+(z-x)^2\}}2$
$x+y+z=k(a^4-b^2c^2+b^4-c^2a^2+c^4-a^2b^2)=\frac k2\{(a^2-bc)^2+(b^2-ca)^2+(c^2-ab)^2\}$
and $x-y=k\{a^4-b^2c^2-(b^4-c^2a^2)\}=k(a^2+b^2+c^2)(a^2-b^2)$
$\implies (x-y)^2+(y-z)^2+(z-x)^2=k^2(a^2+b^2+c^2)^2\{(a^2-b^2)^2+(b^2-c^2)^2+(c^2-a^2)^2\}=k^2(a^2+b^2+c^2)^2\cdot 2(a^4-b^2c^2+b^4-c^2a^2+c^4-a^2b^2)$
$\implies x^3+y^3+z^3-3xyz=k^3\{(a^2+b^2+c^2)(a^4-b^2c^2+b^4-c^2a^2+c^4-a^2b^2)\}^2=k^3(a^6+b^6+c^6-3a^2b^2c^2)^2 $
$$\implies\frac1k=x^3+y^3+z^3-3xyz=k^3(a^6+b^6+c^6-3a^2b^2c^2)^2$$
$$\implies k^2=\frac1{a^6+b^6+c^6-3a^2b^2c^2}$$ as $a^6+b^6+c^6-3a^2b^2c^2=(a^2+b^2+c^2)\frac{\{(bc-a^2)^2+(ca-b^2)^2+(ab-c^2)^2\}}2\ge 0$ for real $a,b,c$
A: Subtract them we get
$$(x^2-y^2)+xz-yz = a^2-b^2 \implies (x-y)(x+y+z) = (a-b)(a+b)$$
$$(y^2-z^2)+yx-zx = b^2-c^2 \implies (y-z)(x+y+z) = (b-c)(b+c)$$
$$(z^2-x^2)+zy-xy = c^2-a^2 \implies (z-x)(x+y+z) = (c-a)(c+a)$$
Now consider the cases $x+y+z=0$ and $x+y+z\neq 0$. I trust you can finish it off from here.
