How to Factorise $x^2 + y^2 + z^2 - 2xy - 2yz - 2zx$ How do you factorise
$x^2 + y^2 + z^2 - 2xy - 2yz - 2zx$ ?
If those minus sings were pluses it would be a lot simpler. I wondered if there was some kind of use of imaginary units which would yield those minus signs.
The factorisation could be done by introducing 3 units, $i, j, k$ such that,
$$
i^2 = j^2 = k^2 = 1 \\
ij = jk = ki = -1
$$
which sort of look like quaternions, but not quite. In this case the factorisation becomes,
$$
x^2 + y^2 + z^2 - 2xy - 2yz - 2zx = (ix + jy + kz)^2.
$$
I didn't know if there was some algebra which obeyed these properties.
 A: As a general principle, you should expect that an inhomogeneous polynomial in $\ge2$ variables, or a homogeneous polynomial in $\ge3$ variables will be irreducible, i.e. hove no further factorization.
An almost-always-valid technique in a case like yours is to set one of the homogeneous variables to $1$ and look at the new inhomogeneous equation. Setting $z=1$, you get
$$x^2+y^2+1-2xy-2x-2y=0\,,$$
which is a nice parabola, axis along the diagonal. If your polynomial had been factorable, the graph would have been two lines.
So my recommendation to you is that you stop and admire the polynomial as it stands. It’s not at all clear to me what any proposed factorization would mean, if the variables were going to be allowed to be in a noncommutative ring: $xy$ would not be $yx$, for instance, and you would have to throw out tools like the Binomial Theorem.
A: Solving the equation
$f(x,y,z)=x^{2} + y^{2} + z^{2} - 2xy - 2yz - 2zx=0$,
compared to $z$ for example, we factorise:
$x^{2} + y^{2} + z^{2} - 2xy - 2yz - 2zx=$
$(x+y-z+2\sqrt{xy})(x+y-z-2\sqrt{xy})$.
