Q:

"Factorizing" a cross-product

Reading further into Newell's Method for computing the normal of an arbitrary polygon, I came across this SO post which states:
$$(P_{1} - P_{0}) \times (P_{2} - P_{0}) \equiv (P_{0} \times P_{1}) + (P_{1} \times P_{2}) + (P_{2} \times P_{0})$$
Where:
$$P_{n>0} = (x_{n} - x_{0}, y_{n} - y_{0})$$
Meaning that each $P_{n}$ represents a vector from two points along the edge of triangle.
Algebraically how is this factor equivalent?

A:

The cross product is distributive over addition. So, just like when you first learned about expanding double brackets:
$$\begin{align}
(P_{1} - P_{0}) \times (P_{2} - P_{0}) &\equiv P_1 \times(P_2 - P_{0}) -P_0 \times (P_{2} - P_{0})
\\ &\equiv P_1 \times P_2 - P_1\times P_0 - P_0\times P_2 + P_0 \times P_0 \tag{1} 
\end{align}$$
We now need some other basic properties of the cross product. For any vectors $u$ and $v$, it is true that:
$$u \times u = 0$$
$$ u \times v = - (v \times u)$$
And so, in $(1)$, the last term equals zero and we can swap vectors around to get rid of the negatives, so that:
$$\begin{align}
(P_{1} - P_{0}) \times (P_{2} - P_{0}) 
 &\equiv P_1 \times P_2 + P_0\times P_1 + P_2\times P_0
\\ &\equiv P_0\times P_1 + P_1 \times P_2 + P_2\times P_0  \tag{2} 
\end{align}$$
as required.

