Solve $x + y + z = xyz$ such that $x , y , z \neq0$ I came across the equation $x+y+z=xyz$ such that $x , y , z \neq 0$.
I set $x=1, y=2, z=3$ but how can i reach formal mathematical solution without " guessing " the answer ? Thank you
 A: If you are looking for positive $$x,y,z$$ then the following image provides a nice geometrical interpretation of your problem: 

Proof
Consider the complex product $$(1+ix)(1+iy)(1+iz)$$ of which the imaginary part is zero, since the sum of the arguments of the complex factors equals: $$\alpha + \beta + \gamma = \pi$$
However, since $$\Im((1+ix)(1+iy)(1+iz)) = (x+y+z) - xyz$$ we have:  $$x+y+z = xyz $$    ∎
A: For:$$A,B,C\in (0,\frac{\pi}{2}),and \,A+B+C=\pi$$
then:$$\tan{A}+\tan{B}+\tan{C}=\tan{A}\tan{B}\tan{C}$$
A: The condition
$$x+y+z=xyz\ne0\tag{1}$$
defines a certain set $S$ in $(x,y,z)$-space. This set is only implicitly described by $(1)$, which means that $(1)$ provides a quick test whether any given point $P=(x,y,z)$ belongs to $S$ or not.
What you want is an explicit description of $S$ in the form of a list of points (if $S$ were finite), or a parametric representation with free variables. Since $(1)$ can be algebraically solved for $z$ in the form
$$z={x+y\over xy-1}$$
we see that above each point $(x,y)\in{\mathbb R}^2$ lies at most one point of $S$. Therefore we immediately arrive at the parametric representation
$$S:\quad (x,y)\mapsto\left(x,y,{x+y\over xy-1}\right)\ .$$
So far this is just formal, and we have to exclude certain parameter points $(x,y)$. These are the points $(x,y)$ on the hyperbola $xy=1$, the points on the $x$- and the $y$ axis, and the points on the line $x+y=0$. All in all we come to the conclusion that $S$ is a union of $8$ disjoint surfaces, each of these lying over some part of the $(x,y)$-plane.
A: you can write $y+z=x(yz-1)$ and if $yz\ne1$ you will get $$x=\frac{y+z}{yz-1}$$
if $yz=1$ we get $y+z=0$ or $y^2+1=0$ which gives us complex solutions.
A: Solve for one variable. As the problem is symmetric, wlog solve for $z$:
$$z = \frac{x+y}{xy-1}.$$
Now you can choose any real values for $x$ and $y$, as long as $xy-1\ne0$. For your choice of $x=1$, $y=2$, you can see it gives the correct result.
If you need a solution in $\mathbb Z$, things get more complicated as the value of $z$ could be a rational for most choices of $x$ and $y$.
A: Let $x = 1$, then $yz = y+z$.
Let $y = z$, then $y\cdot y = y^2 = 2y$.
$\frac{y^2}y = y = 2$.
Using substitution of $x=1$, $y=2$:
$1 + 2 + z = 2z$
$1 + 2 = 2z-z$
$z=3$
$1\times2\times3 = 1 + 2 + 3$
If $x = y = z$, $x$ to $z$ would equal $3^{\frac12}$ .
