Question

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)`?

Answer 1

See below.

Explanation:

`{(x=bz + c y),(y = az + cx),(z = ay + bx):}`

solving for couples we obtain

(1-1) `{(x = ((b + a c) z)/(c^2-1)), (y = ((a + b c) z)/(c^2-1)):}`

(1-2) `{(y = ((a b + c) x)/(a^2-1)), (z = ((b + a c) x)/(a^2-1)):}`

(1-3) `{(z = ((a + b c) y)/(b^2-1)), (x = ((a b + c) y)/(b^2-1)):}`

From (1-1)

`z/(c^2-1)=x/(b+ac) -> z^2/(c^2-1)= (x z)/(b+ac)`

From (1-2)

`x/(a^2-1)=z/(b+ac)->x^2/(c^2-1) = (x z)/(b+ac)`

and also

`z/(c^2-1)=y/(a+bc)->z^2/(c^2-1)=(yz)/(a+bc)`

`y/(b^2-1)=z/(a+bc)->y^2/(b^2-1) = (yz)/(a+bc)`

then follows

`x^2/(a^2-1)=y^2/(b^2-1)=z^2/(c^2-1)`