Solve in $\mathbb R^{3}$ the following system : $\begin{cases}xyz(x^{2}+y^{2})=3\\2xyz(y^{2}+z^{2})=3\\2xyz(x^{2}+z^{2})=5\end{cases}$

Question

I would like to see the solution of the following system equation in $\mathbb R^{3}$

$$\begin{cases}2xyz(x^{2}+y^{2})=6\\2xyz(y^{2}+z^{2})=3\\2xyz(x^{2}+z^{2})=5\end{cases}$$

My try as following :

let $xyz=t$ then $x^{2}+y^{2}=\frac{t^{2}}{(yz)^{2}}+\frac{t^{2}}{(xz)^{2}}$

And : $x^{2}+z^{2}=\frac{t^{2}}{(yz)^{2}}+\frac{t^{2}}{(xy)^{2}}$

$y^{2}+z^{2}=\frac{t^{2}}{(xz)^{2}}+\frac{t^{2}}{(xy)^{2}}$

But I got a difficult system ??

Answer 1

HINT

Use as variables

$$xyz^3\quad xy^3z\quad x^3yz$$

Answer 2

Note that the first equation minus the second gets us $$2xyz(x^2-z^2)=3$$Adding this to the third equation gives$$xyz(x^2)=2$$Likewise, compute $xyz(y^2)$ and $xyz(z^2)$. You can then take their product, and take the fifth root to get the value of $xyz$, after which I'm sure you can finish.

Answer 3

Your approach is similar to the two first answers and works as well.

With $t = xyz$, you get

$$x^2+y^2=3/t$$

$$2y^2+2z^2=3/t$$

$$2x^2+2z^2=5/t$$

So a linear system with variables $x^2, y^2, z^2$ and a parameter $t$.

It is easy solved as

$$x^2 = 2/t$$

$$y^2=1/t$$

$$z^2=1/2t$$

We then simply calculate the product :

$$x^2y^2z^2 = t^2 = \frac{1}{t^3}$$

To get

$$t=1$$