Method of completing squares with 3 variables I want to use the method "completing squares" for this term:
$x^2-2xy +y^2+z^2*a+2xz-2yz$
The result should be $(x-y+z)^2 +(a-1)*z^3$
Is there a "recipe" behind how to do this? Hope someone could help

Aras' answer is really good.  But you might be wondering how one develops intuition and what happens if one doesn't immediately see $x^2 - 2xy + y^2$ is a square. (Part of developing intuition is that this does become obvious.)
This'll get messy. But 
So you have:
$\color {blue}{x^2}-2xy +y^2+z^2*a+2xz-2yz$
And you have an $x^2$ so we'll "complete" x first so take all the x terms and put them nearby.
$\color {blue}{x^2}\color {purple}{-2xy +2xz}+y^2+z^2*a-2yz$
We know that completing the square, the expression will need to start something like this: $x^2 \pm 2*x*something + ....$ so we factor out the $2x$
$\color {blue}{x^2 + 2x}\color {purple}{(z - y)}+y^2+z^2*a-2yz$
Now we know to "complete the square" we must do $x^2 \pm 2x*something \color{green}{+ something^2} \color {red}{- something^2} +...$
So 
$\color {blue}{x^2 + 2x(z - y)} \color{green}{ + (z -y)^2} \color{red}{- (z -y)^2}+y^2+z^2*a-2yz=$
$\color {blue}{(x + (y -z))^2} \color{red}{- (z^2 - 2zy + y^2) }+y^2+z^2*a-2yz$
So let's clean up a little.
$\color {blue}{(x + y -z)^2} \color{red}{- z^2 + 2zy - y^2 }+y^2+z^2*a-2yz$
$\color {blue}{(x + y -z)^2} -z^2  +z^2*a$
So now we need to deal with just the $ ... - z^2 + z^2*a$.  Well we factor out common terms.
$\color {blue}{(x + y -z)^2} +z^2(-1  +a)$
$\color {blue}{(x + y -z)^2} +z^2(a-1)$
