1. ## question about reversing equations

Have any ways to reverse from LHS variables by RHS variables of a group of equations?
For example:

$\displaystyle \begin{matrix}a^2+2bc=x \\ b^2+2ac=y \\ c^2+2ab=z \end{matrix}$

How can I reverse those equations to new equations set likes:

$\displaystyle \begin{matrix}f(x,y,z)=a \\ g(x,y,z)=b \\ h(x,y,z)=c \end{matrix}$

What are those functions look likes?
Thanks

2. Mathematica spits out 8 solutions, but they are extraordinarily long. I'm not even going to attempt to post them here, as they are multi-line solutions. Algebraically, though, there's nothing more complicated than powers and square roots and fractions, and such. I would try to find someone who has Mathematica, and enter the following command:

Simplify[Solve[{a^2 + 2b c == x, b^2 + 2a c == y, c^2 + 2a b == z}, {a, b, c}]].

That should do it.

3. Originally Posted by haedious
Have any ways to reverse from LHS variables by RHS variables of a group of equations?
For example:

$\displaystyle \begin{matrix}a^2+2bc=x \\ b^2+2ac=y \\ c^2+2ab=z \end{matrix}$

How can I reverse those equations to new equations set likes:

$\displaystyle \begin{matrix}f(x,y,z)=a \\ g(x,y,z)=b \\ h(x,y,z)=c \end{matrix}$

What are those functions look likes?
Thanks
Let $\displaystyle \omega = e^{2\pi i/3}$ be a complex cube root of unity. Then

$\displaystyle (a+b+c)^2 = a^2 +b^2 +c^2 +2bc+2ca+2ab = x+y+z,$
$\displaystyle (a+\omega b+\omega^2c)^2 = a^2 +\omega^2b^2 +\omega c^2 +2bc+2\omega^2ca+2\omega ab = x+\omega^2y+\omega z,$
$\displaystyle (a+\omega^2 b+\omega c)^2 = a^2 +\omega b^2 +\omega^2 c^2 +2bc+2\omega ca+2\omega^2 ab = x+\omega y+\omega^2 z.$

Let $\displaystyle X$ be one of the two square roots of $\displaystyle x+y+z$, and similarly let $\displaystyle Y = \bigl(x+\omega y+\omega^2 z\bigr)^{1/2}$ and $\displaystyle Z = \bigl(x+\omega^2 y+\omega z\bigr)^{1/2}.$ Thus there are eight possible choices for the triple of complex numbers $\displaystyle (X,Y,Z).$

Then $\displaystyle a+b+c=X$, $\displaystyle a+\omega^2 b+\omega c = Y$ and $\displaystyle a+\omega b+\omega^2c = Z.$

Since $\displaystyle 1+\omega+\omega^2=0$, it follows that $\displaystyle X+Y+Z = 3a$, $\displaystyle X+\omega Y + \omega^2 Z = 3b$ and $\displaystyle X+\omega^2 Y + \omega Z = 3c.$

That gives the solution in the neat form

$\displaystyle a = \frac13(X+Y+Z),$
$\displaystyle b = \frac13(X+\omega Y + \omega^2 Z),$
$\displaystyle c = \frac13(X+\omega^2 Y + \omega Z).$

Of course, if you write X, Y and Z in terms of x, y and z, it won't look quite so neat. But I reckon that I can claim Opalg: 1, Mathematica: 0.

4. Originally Posted by Opalg
Of course, if you write X, Y and Z in terms of x, y and z, it won't look quite so neat. But I reckon that I can claim Opalg: 1, Mathematica: 0.

5. Originally Posted by Ackbeet
Mathematica spits out 8 solutions, but they are extraordinarily long. I'm not even going to attempt to post them here, as they are multi-line solutions. Algebraically, though, there's nothing more complicated than powers and square roots and fractions, and such. I would try to find someone who has Mathematica, and enter the following command:

Simplify[Solve[{a^2 + 2b c == x, b^2 + 2a c == y, c^2 + 2a b == z}, {a, b, c}]].

That should do it.
I'm not familiar with Mathematica but thank for you help.

6. Originally Posted by Opalg
Let $\displaystyle \omega = e^{2\pi i/3}$ be a complex cube root of unity. Then

$\displaystyle (a+b+c)^2 = a^2 +b^2 +c^2 +2bc+2ca+2ab = x+y+z,$
$\displaystyle (a+\omega b+\omega^2c)^2 = a^2 +\omega^2b^2 +\omega c^2 +2bc+2\omega^2ca+2\omega ab = x+\omega^2y+\omega z,$
$\displaystyle (a+\omega^2 b+\omega c)^2 = a^2 +\omega b^2 +\omega^2 c^2 +2bc+2\omega ca+2\omega^2 ab = x+\omega y+\omega^2 z.$

Let $\displaystyle X$ be one of the two square roots of $\displaystyle x+y+z$, and similarly let $\displaystyle Y = \bigl(x+\omega y+\omega^2 z\bigr)^{1/2}$ and $\displaystyle Z = \bigl(x+\omega^2 y+\omega z\bigr)^{1/2}.$ Thus there are eight possible choices for the triple of complex numbers $\displaystyle (X,Y,Z).$

Then $\displaystyle a+b+c=X$, $\displaystyle a+\omega^2 b+\omega c = Y$ and $\displaystyle a+\omega b+\omega^2c = Z.$

Since $\displaystyle 1+\omega+\omega^2=0$, it follows that $\displaystyle X+Y+Z = 3a$, $\displaystyle X+\omega Y + \omega^2 Z = 3b$ and $\displaystyle X+\omega^2 Y + \omega Z = 3c.$

That gives the solution in the neat form

$\displaystyle a = \frac13(X+Y+Z),$
$\displaystyle b = \frac13(X+\omega Y + \omega^2 Z),$
$\displaystyle c = \frac13(X+\omega^2 Y + \omega Z).$

Of course, if you write X, Y and Z in terms of x, y and z, it won't look quite so neat. But I reckon that I can claim Opalg: 1, Mathematica: 0.
It's great for having steps to the solution even I don't understand completely. And I never thought that how apply cube root of unity to solve this issue so I've difficult to digest these steps but still thanks.

I treats above question similar as just a square root to a number consisted with {x,y,z}.