# Given x=bz+cy;" " y=az+cx and z=ay+bx how will you prove x^2/(1-a^2)=y^2/(1-b^2)=z^2/(1-c^2)?

Dec 19, 2016

See below.

#### Explanation:

$\left\{\begin{matrix}x = b z + c y \\ y = a z + c x \\ z = a y + b x\end{matrix}\right.$

solving for couples we obtain

(1-1) $\left\{\begin{matrix}x = \frac{\left(b + a c\right) z}{{c}^{2} - 1} \\ y = \frac{\left(a + b c\right) z}{{c}^{2} - 1}\end{matrix}\right.$

(1-2) $\left\{\begin{matrix}y = \frac{\left(a b + c\right) x}{{a}^{2} - 1} \\ z = \frac{\left(b + a c\right) x}{{a}^{2} - 1}\end{matrix}\right.$

(1-3) $\left\{\begin{matrix}z = \frac{\left(a + b c\right) y}{{b}^{2} - 1} \\ x = \frac{\left(a b + c\right) y}{{b}^{2} - 1}\end{matrix}\right.$

From (1-1)

$\frac{z}{{c}^{2} - 1} = \frac{x}{b + a c} \to {z}^{2} / \left({c}^{2} - 1\right) = \frac{x z}{b + a c}$

From (1-2)

$\frac{x}{{a}^{2} - 1} = \frac{z}{b + a c} \to {x}^{2} / \left({c}^{2} - 1\right) = \frac{x z}{b + a c}$

and also

$\frac{z}{{c}^{2} - 1} = \frac{y}{a + b c} \to {z}^{2} / \left({c}^{2} - 1\right) = \frac{y z}{a + b c}$

$\frac{y}{{b}^{2} - 1} = \frac{z}{a + b c} \to {y}^{2} / \left({b}^{2} - 1\right) = \frac{y z}{a + b c}$

then follows

${x}^{2} / \left({a}^{2} - 1\right) = {y}^{2} / \left({b}^{2} - 1\right) = {z}^{2} / \left({c}^{2} - 1\right)$