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

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

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

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

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:

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

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

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

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

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

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

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

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

That gives the solution in the neat form

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

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 $\omega = e^{2\pi i/3}$ be a complex cube root of unity. Then

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

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

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

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

That gives the solution in the neat form

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

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.