<s>[INST] 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:$x=\pm\dfrac{a^4-b^2c^2}{\sqrt {a^6+b^6+c^6-3a^2b^2c^2}}\,$, $y=\pm\dfrac{b^4-a^2c^2}{\sqrt {a^6+b^6+c^6-3a^2b^2c^2}}\,$ ,$z=\pm\dfrac{c^4-b^2a^2}{\sqrt {a^6+b^6+c^6-3a^2b^2c^2}}$
 [/INST] 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.
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