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Given the system: $$ \begin{cases} x^2y^2+x^2z^2=axyz & \\ y^2z^2+y^2x^2=bxyz &\\ z^2x^2+z^2y^2=cxyz \end{cases} $$ The solution could be gotten in a very tedious way. Is it possible to solve it considering some symmetry property of the system if any? Thanks in advance.

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Factor, factor, factor. –  John Jul 18 at 22:24

3 Answers 3

up vote 8 down vote accepted

Divide all three equations by $x y z$, and observe that the resulting system of equations has in effect only three variables. This is readily solved, and one can then map these back to $\{x,y,z\}$ rather easily.

Edit: As noted by Yeldarbskich, the division by $x y z$ is only valid if all are nonzero. Consequently there will also be three families of solutions corresponding to (at least) two of $\{x,y,z\}$ being zero.

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From the start, the system has three variables, right? Can you please elaborate a bit ? –  lab bhattacharjee Jul 18 at 16:45
    
I was being deliberately a bit glib. What I meant more specifically is that after dividing through by $x y z$, the three equations are functions of $\{xy/z,y z/x,z x/y\}.$ –  Semiclassical Jul 18 at 16:46

If is clear we cannot have some $a,b,c$ positive while the rest negative. Otherwise, $xyz = 0$ and we only have trivial solution.

For simplicity, I will assume $a, b, c > 0$.

If any one of $x,y,z$ vanishes, say $x = 0$, it is clear the equations implies $y z = 0$. This means if one of $x,y,z$ vanish, at least two of $x,y,z$ vanish. If two of $x,y,z$ vanish, then the third one is arbitrary.

To eliminate this sort of trivial solutions, we will first assume $x, y, z > 0$. If we divide everything by $x^2 y^2 z^2$, we have $$\begin{cases} \frac{1}{z^2} + \frac{1}{y^2} &= \frac{a}{xyz}\\ \frac{1}{x^2} + \frac{1}{z^2} &= \frac{b}{xyz}\\ \frac{1}{y^2} + \frac{1}{x^2} &= \frac{c}{xyz} \end{cases} \quad\iff\quad \begin{cases} \frac{1}{x^2} &= \frac{b+c-a}{2xyz}\\ \frac{1}{y^2} &= \frac{a+c-b}{2xyz}\\ \frac{1}{z^2} &= \frac{a+b-c}{2xyz} \end{cases} \tag{*1} $$ This implies $$\begin{align} \frac{1}{(xyz)^2} &= \left(\frac{b+c-a}{2}\right)\left(\frac{a+c-b}{2}\right)\left(\frac{a+b-c}{2}\right)\frac{1}{(xyz)^3}\\ \iff\quad\quad xyz &= \left(\frac{b+c-a}{2}\right)\left(\frac{a+c-b}{2}\right)\left(\frac{a+b-c}{2}\right) \end{align} $$ Substitute this back into $(*1)$ immediately give us

$$\begin{cases} x &= x_0 \stackrel{def}{=} \sqrt{\left(\frac{a+c-b}{2}\right)\left(\frac{a+b-c}{2}\right)}\\ y &= y_0 \stackrel{def}{=} \sqrt{\left(\frac{b+c-a}{2}\right)\left(\frac{a+b-c}{2}\right)}\\ z &= z_0 \stackrel{def}{=} \sqrt{\left(\frac{b+c-a}{2}\right)\left(\frac{a+c-b}{2}\right)} \end{cases}\tag{*2}$$

What happens if some or all of the $x,y,z$ are negative? If one look back the original set of equations, the LHS is positive. Under the assume that $a, b, c > 0$, we find $xyz$ is positive. This means in general, the set of non-trivial solutions of the original equation is given by that in $(*2)$ with an even number of $x,y,z$ flipped sign. i.e. $$(x,y,z) = (x_0,y_0,z_0), (x_0,-y_0,-z_0), (-x_0,y_0,-z_0) \text{ or } (-x_0,-y_0,z_0)$$

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I just want to add that Semiclassical's answer above has given you the right idea, but one should be careful to note that the possibility of any of x,y, and z could be zero, which would not work for that method. You should go back and check for solutions for combinations of x,y, and z being zero.

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In particular, Semiclassical's method won't give you the trivial solution x=y=z=0. –  Yeldarbskich Jul 18 at 16:53
    
Good point. Mea culpa. –  Semiclassical Jul 18 at 16:55
    
It's not that bad actually. I just looked and if any are zero then you must have the trivial solution, but a solution nonetheless. –  Yeldarbskich Jul 18 at 16:56
    
Actually, if any two of $x$, $y$ and $z$ are zero, then the equations are trivially satisfied regardless of the value of third variable (or of the coefficients $a$, $b$ and $c$). So there are three families of trivial solutions, which correspond to the three coordinate axes in $(x,y,z)$ space, and which intersect at the point $x=y=z=0$. –  Ilmari Karonen Jul 18 at 18:44
    
"Good point. Mea culpa." –  Yeldarbskich Jul 18 at 20:09

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