Question

Solve the following equations :
$x(y+z)=a^2,y(z+x)=b^2,z(x+y)=c^2.$

Answer 1

The given system of equations can be written as
$xy+xz=a^2$
$yz+yz=b^2$
$zx+zy=c^2$
Adding, $xy+yz+zx=\frac{1}{2}(a^2+b^2+c^2)$
Then from $\left(1\right)$ and $\left(4\right)$, $yz=\frac{1}{2}(b^2+c^2+a^2)$
Similarly, $zx=\frac{c^2+a^2-b^2}{2}$ and $xy=\frac{a^2+b^2-c^2}{2}$
Multiplying these,
${\left(xyz\right)}^2=\frac{(b^2+c^2+a^2)(c^2+a^2+b^2)(a^2+b^2+c^2)}{8}$
$\therefore xyz=\pm \sqrt{\frac{1}{8}(b^2+c^2-a^2)(c^2+a^2-b^2)(a^2+b^2-c^2)}$
Now we easily get
$x=\pm \sqrt{\left\{\frac{(c^2+a^2-b^2)(a^2+b^2-c^2)}{2(b^2+c^2-a^2)}\right\}}$
$y=\pm \sqrt{\left\{\frac{(a^2+b^2-c^2)(b^2+c^2-a^2)}{2(c^2+a^2-b^2)}\right\}}$
$z=\pm \sqrt{\left\{\frac{(c^2+a^2-b^2)(b^2+c^2-a^2)}{2(a^2+b^2-c^2)}\right\}}$.