Question

Eliminate $x,y,z$ between the equations $\frac{y}{z}-\frac{z}{y}=a,\frac{z}{x}-\frac{x}{z}=b,\frac{x}{y}-\frac{y}{x}=c$.

Answer 1

We have $a+b+c=\frac{x(y^2-z^2)+y(z^2-x^2)+z(x^2-y^2)}{xyz}$

$=\frac{x(y^2-z^2)-y(y^2-z^2+x^2-y^2)+z(x^2-y^2)}{xyz}$
$=\frac{(y^2-z^2)(x-y)+(x^2-y^2)(z-y)}{xyz}$
$=\frac{(y-z)(y+z)(x-y)+(x-y)(x+y)(z-y)}{xyz}$
$=\frac{(y-z)(x-y)(y+z-x-y)}{xyz}$
$=\frac{(y-z)(x-y)(z-x)}{xyz}$
If we change the sign of x, the signs of $b$ and $c$ are changed, while the sign of $a$ remains unaltered
$\therefore a-b-c=\frac{(y-z)(x+y)(z+x)}{xyz}$
Similarly, we change the sign of y and z, we get
$b-c-a=\frac{(y+z)(x+y)(z-x)}{xyz}$
$c-a-b=\frac{(y+z)(x-y)(z+x)}{xyz}$
Multiply the four equations, we get
$(a+b+c)(a-b-c)(b-c-a)(c-a-b)=\frac{(y^2-z^2{)}^2(z^2-x^2{)}^2(x^2-y^2{)}^2}{x^4{y}^4{z}^4}$
$=-{((\frac{y}{z}-\frac{z}{y})\times (\frac{z}{x}-\frac{x}{z})\times (\frac{x}{y}-\frac{y}{x}))}^2$
$=-a^2{b}^2{c}^2.$
Therefore, the eliminated equation is
$2b^2{c}^2+2a^2{b}^2+2c^2{a}^2-a^4-b^4-c^4+a^2{b}^2{c}^2=0$