Question 252008

if {{{x=cy+bz}}}
   {{{y=az+cx}}}
   {{{z=bx+ay}}}
 
Then prove that {{{x^2/(1-a^2)=y^2/(1-b^2)=z^2/(1-c^2)}}}
<pre><font size = 4 color = "indigo"><b>                
Solve the second for c:
{{{y=az+cx}}}
{{{y-az=cx}}}
{{{(y-az)/x=c}}}
Use the third equation to substitute for z
{{{(y-a(bx+ay))/x}=c}}
{{{(y-abx-a^2y)/x=c}}}

Use this and the third original equation to substitute
for both c and z in the first equation:


{{{x=cy+bz}}}
{{{x=((y-abx-a^2y)/x)y+b(bx+ay)}}}
Multiply thru by x:
{{{x^2=(y-abx-a^2y)y+bx(bx+ay)}}}
{{{x^2=y^2-abxy-a^2y^2+b^2x^2+abxy}}}
{{{x^2=y^2-cross(abxy)-a^2y^2+b^2x^2+cross(abxy)}}}
{{{x^2=y^2-a^2y^2+b^2x^2}}}
{{{x^2-b^2x^2=y^2-a^2y^2}}}
{{{x^2(1-b^2)=y^2(1-a^2)}}}

Divide both sides by {{{(1-a^2)(1-b^2)}}}

{{{(x^2(1-b^2))/((1-a^2)(1-b^2))
=(y^2(1-a^2))/((1-a^2)(1-b^2))
}}}

{{{(x^2cross((1-b^2)))/((1-a^2)cross((1-b^2)))
=(y^2cross((1-a^2)))/(cross((1-a^2))(1-b^2))
}}}

{{{x^2/(1-a^2)=y^2/(1-b^2)}}} 

To prove that one of these also equals {{{z^2/(1-c^2)}}}
is exactly similar to the above.

Edwin</pre>