Question

Solve the following equations :
$x^2-yz=a^2,y^2-zx=b^2,z^2-xy=c^2.$

Answer 1

Multiplying the given equations by $y,z,x$ respectively and adding, we get.
$c^{2}x+a^{2}y+b^{2}z=0$.
Again multiplying the given equations by $z,x,y$ respectively and adding, we get.
$b^{2}x+c^{2}y+a^{2}z=0$.
Solving $\left(1\right)$ and $\left(2\right)$ by cross-multiplication,
$\frac{x}{a^4-b^2{c}^2}=\frac{y}{b^4-c^2{a}^2}=\frac{z}{c^4-a^2{b}^2}=k,$ say
Substituting in any one of the given equations, for $x,y,z$ in terms of $k$ from $\left(3\right),$ we get
$k^2(a^6+b^6+c^6-3a^2{b}^2{c}^2)=1$
$\therefore \frac{x}{a^4-b^2{c}^2}=\frac{y}{b^4-c^2{a}^2}=\frac{z}{c^4-a^2{b}^2}$
$=\pm \frac{1}{\surd [a^6+b^6+c^6-3a^2{b}^2{c}^2]}$