Question

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

Answer 1

$x^2-yz=a^2$ .......(i)

$y^2-zx=b^2$ ........(ii)

$z^2-xy=c^2$ ........(iii)

Multiplying (i) by $y$, (ii) by $z$, (iii) by $x$ and adding

$\Rightarrow c^{2}x+a^{2}y+b^{2}z=0$ .........(1)

Now multiplying (i) by $z$, (ii) by $x$, (iii) by $y$ and adding

$\Rightarrow b^{2}x+c^{2}y+a^{2}z=0$ ............(2)

Solving (1) and (2) by cross multiplication

$\frac{x}{a^4-b^2{c}^2}=\frac{y}{b^4-c^2{a}^2}=\frac{z}{c^4-a^2{b}^2}=k\Rightarrow x=k(a^4-b^2{c}^2),y=k(b^4-c^2{a}^2),z=k(c^4-a^2{b}^2).....\left(iv\right)$

Substituting $x,y,z$ in (i)

${\left(k\right(a^4-b^2{c}^2\left)\right)}^2-k^2(b^4-c^2{a}^2)(c^4-a^2{b}^2)=a^2\Rightarrow k^2\left(\right({a^4-b^2{c}^2)}^2-(b^4-c^2{a}^2)(c^4-a^2{b}^2))=a^2\Rightarrow k^2(a^8{+b}^4{c}^4-2a^4{b}^2{c}^2-b^4{c}^4+a^2{b}^6+a^2{c}^6-2a^4{b}^2{c}^2)=a^2\Rightarrow k^2=\frac{a^2}{a^8+a^2{b}^6+a^2{c}^6-3a^4{b}^2{c}^2}\Rightarrow k^2=\frac{1}{a^6+b^6+c^6-3a^2{b}^2{c}^2}\Rightarrow k=\pm \frac{1}{\sqrt{a^6+b^6+c^6-3a^2{b}^2{c}^2}}$

Substituting $k$ in (iv)
$\Rightarrow x=\pm \frac{a^4-b^2{c}^2}{\sqrt{a^6+b^6+c^6-3a^2{b}^2{c}^2}}\Rightarrow y=\pm \frac{b^4-c^2{a}^2}{\sqrt{a^6+b^6+c^6-3a^2{b}^2{c}^2}}\Rightarrow z=\pm \frac{c^4-a^2{b}^2}{\sqrt{a^6+b^6+c^6-3a^2{b}^2{c}^2}}$