Question

Solve the following equations :
$y+z-x=\frac{xyz}{a^2},z+x-y=\frac{xyz}{b^2},x+y-z=\frac{xyz}{c^2}.$

Answer 1

One ovious solution is $x=0,y=0,z=0$.
Adding the given equations pairwise, we get
$2z=xyz(\frac{1}{a^2}+\frac{1}{b^2})$ etc.
Hence $\frac{2}{xy}=\frac{1}{a^2}+\frac{1}{b^2}$.
Similarly,$\frac{2}{yz}=\frac{1}{b^2}+\frac{1}{c^2},\frac{2}{zx}=\frac{1}{a^2}+\frac{1}{c^2}$.
Multiplying these, we obtain
$x^2{y}^2{z}^2=\frac{8a^2{b}^2{c}^2}{(a^2+b^2)(b^2+c^2)(c^2+a^2)}$
Hence $xyz=\pm \frac{2\sqrt{2}.a^2{b}^2{c}^2}{\surd \left[\right(a^2+b^2\left)\right(b^2+c^2\left)\right(c^2+a^2\left)\right]}$
Now using the equality $xy=\frac{2a^2{b}^2}{a^2+b^2}$, we find
$z=\pm \frac{\sqrt{2}.c^2\sqrt{(a^2+b^2)}}{\surd \left[\right(b^2+c^2\left)\right(c^2+a^2\left)\right]}$
Similarly expressions for value of $x$ and $y$ can be found.